一、
求逆元-费马小定理
1 //amodm 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 using namespace std; 6 typedef long long lol; 7 8 int n,p; 9 10 lol qpow(lol x) 11 { 12 int y=p-2; 13 lol ans=1; 14 while (y) { 15 if (y&1) ans*=x,ans%=p; 16 x*=x;x%=p;y/=2; 17 } 18 return ans; 19 } 20 21 inline void write(lol x) 22 { 23 if(x<0) putchar(‘-‘),x=-x; 24 if(x>9) write(x/10); 25 putchar(x%10+‘0‘); 26 } 27 28 int main() 29 { 30 scanf("%d%d",&n,&p); 31 for (int i=1;i<=n;i++) { 32 write(qpow(i)%p); 33 putchar(‘\n‘); 34 } 35 return 0; 36 }
二、
树状数组
1 //BIT 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 using namespace std; 6 typedef long long lol; 7 8 9 10 lol n,m; 11 lol f[500010]; 12 13 14 15 lol lowbit(lol x) 16 { 17 return x&-x; 18 } 19 20 21 22 void add(lol x,lol k) 23 { 24 for (int i=x;i<=n;i+=lowbit(i)) f[i]+=k; 25 } 26 27 28 29 lol getsum(lol x) 30 { 31 lol ans=0; 32 for (int i=x;i;i-=lowbit(i)) ans+=f[i]; 33 return ans; 34 } 35 36 37 38 int main() 39 { 40 cin>>n>>m;lol a; 41 for (int i=1;i<=n;i++) scanf("%lld",&a),add(i,a); 42 lol flag,x,y,k; 43 for (int i=1;i<=m;i++) { 44 scanf("%lld%lld",&flag,&x); 45 if (flag==1) { 46 scanf("%lld",&k); 47 add(x,k); 48 } 49 if (flag==2) { 50 scanf("%lld",&y); 51 printf("%lld\n",getsum(y)-getsum(x-1)); 52 } 53 } 54 return 0; 55 }
三、
最大流-Dinic
1 //dinic 2 #include <queue> 3 #include <cstdio> 4 #include <cstring> 5 #include <iostream> 6 using namespace std; 7 8 9 10 int cnt=-1,depth[10010],cur[10010],s,t,n,m; 11 int next[200010],head[10010],v[200010],w[200010]; 12 13 14 15 int dfs(int u,int flow) 16 { 17 if (u==t) return flow; 18 for (int& i=cur[u];i!=-1;i=next[i]) { 19 if ((w[i]!=0)&&(depth[v[i]]==depth[u]+1)) { 20 int di=dfs(v[i],min(flow,w[i])); 21 if (di>0) {w[i]-=di;w[i^1]+=di;return di;} 22 } 23 } 24 return 0; 25 } 26 27 28 29 bool bfs() 30 { 31 queue <int> q; 32 memset(depth,0,sizeof(depth)); 33 while (!q.empty()) q.pop(); 34 q.push(s);depth[s]=1; 35 do { 36 int u=q.front();q.pop(); 37 for (int i=head[u];i!=-1;i=next[i]) { 38 if ((w[i]>0)&&(depth[v[i]]==0)) { 39 q.push(v[i]);depth[v[i]]=depth[u]+1; 40 } 41 } 42 }while(!q.empty()); 43 if (depth[t]>0) return 1; 44 return 0; 45 } 46 47 48 49 int dinic() 50 { 51 int ans=0; 52 while (bfs()) { 53 for (int i=1;i<=n;i++) cur[i]=head[i]; 54 while (int d=dfs(s,3e8)) ans+=d; 55 } 56 return ans; 57 } 58 59 60 61 int main() 62 { 63 int a,b,c; 64 cin>>n>>m>>s>>t; 65 memset(head,-1,sizeof(head)); 66 memset(next,-1,sizeof(next)); 67 while (m--) { 68 scanf("%d%d%d",&a,&b,&c); 69 cnt++;v[cnt]=b;w[cnt]=c; 70 next[cnt]=head[a];head[a]=cnt; 71 cnt++;v[cnt]=a;w[cnt]=0; 72 next[cnt]=head[b];head[b]=cnt; 73 } 74 cout<<dinic()<<endl; 75 return 0; 76 }
四、
二分图-匈牙利算法
1 //eft_xyl 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 using namespace std; 6 7 8 9 int n,m,e,girl[1005]; 10 bool used[1005],line[1005][1005]; 11 12 13 14 bool find(int x) 15 { 16 for (int j=1;j<=m;j++) 17 if (line[x][j]&&!used[j]) { 18 used[j]=1; 19 if (!girl[j]||find(girl[j])) { 20 girl[j]=x;return 1; 21 } 22 } 23 return 0; 24 } 25 26 27 28 int main() 29 { 30 int x,y; 31 cin>>n>>m>>e; 32 for (int i=1;i<=e;i++) 33 scanf("%d%d",&x,&y),line[x][y]=1; 34 e=0; 35 for (int i=1;i<=n;i++) { 36 if (find(i)) e++; 37 memset(used,0,sizeof(used)); 38 } 39 cout<<e<<endl; 40 return 0; 41 }
五、
扩展欧几里得
1 //exgcd 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 using namespace std; 6 7 8 9 void exgcd(int a,int b,int &x,int &y) 10 { 11 if (!b) x=1,y=0; 12 else { 13 exgcd(b,a%b,y,x); 14 y-=x*(a/b); 15 } 16 } 17 18 19 20 int main() 21 { 22 int a,b,x,y; 23 cin>>a>>b; 24 exgcd(a,b,x,y); 25 cout<<(x%b+b)%b<<endl; 26 return 0; 27 }
六、
假的字符串Hash,其实是map
1 //jia de hash,zhe shi map! 2 #include <map> 3 #include <cmath> 4 #include <cstdio> 5 #include <cstring> 6 #include <iostream> 7 using namespace std; 8 9 10 11 int n; 12 string s; 13 map <string,int> mp; 14 15 16 17 int main() 18 { 19 cin>>n;int ans=0; 20 for (int i=1;i<=n;i++) { 21 cin>>s; 22 if (!mp.count(s)) ans++,mp[s]=1; 23 } 24 cout<<ans<<endl; 25 return 0; 26 }
七、
克鲁斯卡尔
1 //kruskal 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 #include <algorithm> 6 using namespace std; 7 8 9 10 int n,m; 11 int fa[5001]; 12 13 14 15 struct ed { 16 int f,t,w; 17 }e[400010]; 18 19 20 21 bool cmp(ed a,ed b) 22 { 23 return a.w<b.w; 24 } 25 26 27 28 void init() 29 { 30 for (int i=1;i<=n;i++) fa[i]=i; 31 } 32 33 34 35 int find(int a) 36 { 37 if (a==fa[a]) return a; 38 return fa[a]=find(fa[a]); 39 } 40 41 42 43 void unionset(int a,int b) 44 { 45 int u=find(a),v=find(b); 46 if (u!=v) fa[u]=v; 47 } 48 49 50 51 bool same(int u,int v) 52 { 53 return find(u)==find(v); 54 } 55 56 57 58 void kruskal() 59 { 60 int res=0; 61 init();int cnt=0; 62 sort(e+1,e+1+m,cmp); 63 for (int i=1;i<=m;i++) { 64 if (!same(e[i].f,e[i].t)) { 65 res+=e[i].w; 66 unionset(e[i].f,e[i].t); 67 ++cnt; 68 } 69 if (cnt==n-1) break; 70 } 71 cout<<res<<endl; 72 } 73 74 75 76 int main() 77 { 78 cin>>n>>m;int x,y,w; 79 for (int i=1;i<=m;i++) { 80 scanf("%d%d%d",&x,&y,&w); 81 e[i].f=x;e[i].t=y;e[i].w=w; 82 } 83 kruskal(); 84 return 0; 85 }
八、
LCA-倍增
1 //lca-ST 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 using namespace std; 6 7 8 9 int n,m,ss; 10 int dep[500010]; 11 int dpt[500010][30]; 12 13 14 15 int cnt; 16 int o[1000010]; 17 int s[1000010][2]; 18 //lsqxx!!! 19 void jia(int x,int y) 20 { 21 s[++cnt][1]=o[x]; 22 s[cnt][0]=y;o[x]=cnt; 23 } 24 25 26 27 void dfs(int x,int d) 28 { 29 dep[x]=d; 30 for (int i=o[x];i;i=s[i][1]) { 31 int y=s[i][0]; 32 if (!dep[y]) { 33 dpt[y][0]=x; 34 dfs(y,d+1); 35 } 36 } 37 } 38 39 40 41 void init() 42 { 43 for (int j=1;j<=19;j++) 44 for (int i=1;i<=n;i++) 45 dpt[i][j]=dpt[dpt[i][j-1]][j-1]; 46 } 47 48 49 50 int lca(int a,int b) 51 { 52 if (dep[a]<dep[b]) swap(a,b); 53 for (int j=19;j>=0;j--) 54 if (dep[a]-(1<<j)>=dep[b]) a=dpt[a][j]; 55 if (a!=b) { 56 for (int j=19;j>=0;j--) 57 if (dpt[a][j]!=dpt[b][j]) 58 a=dpt[a][j],b=dpt[b][j]; 59 a=dpt[a][0]; 60 } 61 return a; 62 } 63 64 65 66 int main() 67 { 68 int x,y; 69 cin>>n>>m>>ss; 70 for (int i=1;i<n;i++) { 71 scanf("%d%d",&x,&y); 72 jia(x,y);jia(y,x); 73 } 74 dfs(ss,1);init(); 75 for (int i=1;i<=m;i++) { 76 scanf("%d%d",&x,&y); 77 printf("%d\n",lca(x,y)); 78 } 79 return 0; 80 }
九、
构造最长公共子序列LCS
1 //只能打印一个最长公共子序列 2 #include <iostream> 3 using namespace std; 4 5 const int X = 100, Y = 100; //串的最大长度 6 char result[X+1]; //用于保存结果 7 int count=0; //用于保存公共最长公共子串的个数 8 9 10 11 /*功能：计算最优值 12 *参数： 13 * x:字符串x 14 * y:字符串y 15 * b:标志数组 16 * xlen:字符串x的长度 17 * ylen:字符串y的长度 18 *返回值：最长公共子序列的长度 19 * 20 */ 21 int Lcs_Length(string x, string y, int b[][Y+1],int xlen,int ylen) 22 { 23 int i = 0; 24 int j = 0; 25 26 27 28 int c[X+1][Y+1]; 29 for (i = 0; i<=xlen; i++) 30 { 31 c[i][0]=0; 32 } 33 for (i = 0; i <= ylen; i++ ) 34 { 35 c[0][i]=0; 36 } 37 for (i = 1; i <= xlen; i++) 38 { 39 40 for (j = 1; j <= ylen; j++) 41 { 42 if (x[i - 1] == y[j - 1]) 43 { 44 c[i][j] = c[i-1][j-1]+1; 45 b[i][j] = 1; 46 } 47 else 48 if (c[i-1][j] > c[i][j-1]) 49 { 50 c[i][j] = c[i-1][j]; 51 b[i][j] = 2; 52 } 53 else 54 if(c[i-1][j] <= c[i][j-1]) 55 { 56 c[i][j] = c[i][j-1]; 57 b[i][j] = 3; 58 } 59 60 } 61 } 62 cout << "计算最优值效果图如下所示：" << endl; 63 for(i = 1; i <= xlen; i++) 64 { 65 for(j = 1; j < ylen; j++) 66 { 67 cout << c[i][j] << " "; 68 } 69 cout << endl; 70 } 71 return c[xlen][ylen]; 72 } 73 74 void Display_Lcs(int i, int j, string x, int b[][Y+1],int current_Len) 75 { 76 if (i ==0 || j==0) 77 { 78 return; 79 } 80 if(b[i][j]== 1) 81 { 82 current_Len--; 83 result[current_Len]=x[i- 1]; 84 Display_Lcs(i-1, j-1, x, b, current_Len); 85 } 86 else 87 { 88 if(b[i][j] == 2) 89 { 90 Display_Lcs(i-1, j, x, b, current_Len); 91 } 92 else 93 { 94 if(b[i][j]==3) 95 { 96 Display_Lcs(i, j-1, x, b, current_Len); 97 } 98 else 99 { 100 Display_Lcs(i-1,j,x,b, current_Len); 101 } 102 } 103 } 104 } 105 106 int main(int argc, char* argv[]) 107 { 108 string x = "ABCBDAB"; 109 string y = "BDCABA"; 110 int xlen = x.length(); 111 int ylen = y.length(); 112 113 114 115 int b[X + 1][Y + 1]; 116 117 118 119 int lcs_max_len = Lcs_Length( x, y, b, xlen,ylen ); 120 cout << lcs_max_len << endl; 121 122 123 124 Display_Lcs( xlen, ylen, x, b, lcs_max_len ); 125 126 //打印结果如下所示 127 for(int i = 0; i < lcs_max_len; i++) 128 { 129 cout << result[i]; 130 } 131 cout << endl; 132 return 0; 133 }
十、
归并排序求逆序对
1 //msort_nxd 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 using namespace std; 6 7 8 9 int ans,n,a[100010],r[100010]; 10 11 12 13 void msort_nxd(int s,int t) 14 { 15 if (s==t) return; 16 int mid=s+t>>1; 17 msort_nxd(s,mid); 18 msort_nxd(mid+1,t); 19 int i=s,j=mid+1,k=s; 20 while (i<=mid&&j<=t) { 21 if (a[i]<=a[j]) { 22 r[k]=a[i];i++;k++; 23 } 24 else { 25 r[k]=a[j];j++;k++; 26 ans+=mid-i+1; 27 } 28 } 29 while (i<=mid) r[k++]=a[i++]; 30 while (j<=t) r[k++]=a[j++]; 31 for (int q=s;q<=t;q++) a[q]=r[q]; 32 } 33 34 35 36 int main() 37 { 38 cin>>n; 39 for (int i=1;i<=n;i++) { 40 scanf("%d",&a[i]); 41 } 42 msort_nxd(1,n); 43 cout<<ans<<endl; 44 return 0; 45 }
十一、
线段树
1 //segment 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 #define ll(x) ((x)<<1) 6 #define rr(x) ((x)<<1|1) 7 using namespace std; 8 typedef long long lol; 9 10 11 12 lol n,m,a[100010]; 13 lol sgm[100010*4],lazy[100010*4]; 14 15 16 17 void shang(lol r) 18 { 19 sgm[r]=sgm[ll(r)]+sgm[rr(r)]; 20 } 21 22 23 24 void xia(lol r,lol z,lol y) 25 { 26 int k=z+y>>1; 27 lazy[ll(r)]+=lazy[r]; 28 lazy[rr(r)]+=lazy[r]; 29 sgm[ll(r)]+=(k-z+1)*lazy[r]; 30 sgm[rr(r)]+=(y-k)*lazy[r]; 31 lazy[r]=0; 32 } 33 34 35 36 void js(lol r,lol z,lol y) 37 { 38 if (z==y) { 39 sgm[r]=a[z]; 40 return; 41 } 42 int k=z+y>>1; 43 js(ll(r),z,k); 44 js(rr(r),k+1,y); 45 shang(r); 46 } 47 48 49 50 void qjxg(lol r,lol z,lol y,lol zz,lol yy,lol v) 51 { 52 if (z>yy||y<zz) return; 53 if (z>=zz&&y<=yy) { 54 lazy[r]+=v; 55 sgm[r]+=(y-z+1)*v; 56 return; 57 } 58 if (lazy[r]) xia(r,z,y); 59 int k=z+y>>1; 60 if (z<=k) qjxg(ll(r),z,k,zz,yy,v); 61 if (y>k) qjxg(rr(r),k+1,y,zz,yy,v); 62 shang(r); 63 } 64 65 66 67 lol cx(lol r,lol z,lol y,lol zz,lol yy) 68 { 69 if (z>yy||y<zz) return 0; 70 if (z>=zz&&y<=yy) return sgm[r]; 71 int k=z+y>>1; 72 if (lazy[r]) xia(r,z,y); 73 return cx(ll(r),z,k,zz,yy)+cx(rr(r),k+1,y,zz,yy); 74 } 75 76 77 78 int main() 79 { 80 cin>>n>>m; 81 for (int i=1;i<=n;i++) { 82 scanf("%lld",&a[i]); 83 } 84 js(1,1,n); 85 lol flag,x,y,k; 86 for (int i=1;i<=m;i++) { 87 scanf("%lld",&flag); 88 if (flag==1) { 89 scanf("%lld%lld%lld",&x,&y,&k); 90 qjxg(1,1,n,x,y,k); 91 } 92 else { 93 scanf("%lld%lld",&x,&y); 94 cout<<cx(1,1,n,x,y)<<endl; 95 } 96 } 97 return 0; 98 }
十二、
SPFA
1 //SPFA 2 #include <queue> 3 #include <cstdio> 4 #include <cstring> 5 #include <iostream> 6 using namespace std; 7 8 9 10 int n,m,st,d[10010]; 11 12 13 14 int cnt; 15 int o[500010]; 16 int s[500010][3]; 17 //lsqxx!!! 18 void jia(int x,int y,int c) 19 { 20 s[++cnt][1]=o[x]; 21 s[cnt][0]=y; 22 s[cnt][2]=c;o[x]=cnt; 23 } 24 25 26 27 queue <int> q; 28 bool v[10010]; 29 30 31 32 void SPFA() 33 { 34 for (int i=1;i<=n;i++) 35 d[i]=2147483647; 36 v[st]=1;q.push(st);d[st]=0; 37 while (!q.empty()) { 38 int x=q.front(); 39 for (int i=o[x];i;i=s[i][1]) { 40 int y=s[i][0]; 41 if (d[y]>d[x]+s[i][2]) { 42 d[y]=d[x]+s[i][2]; 43 if (!v[y]) { 44 v[y]=1; 45 q.push(y); 46 } 47 } 48 } 49 q.pop();v[x]=0; 50 } 51 } 52 53 54 55 int main() 56 { 57 int x,y,c; 58 cin>>n>>m>>st; 59 for (int i=1;i<=m;i++) { 60 scanf("%d%d%d",&x,&y,&c); 61 jia(x,y,c); 62 } 63 SPFA(); 64 for (int i=1;i<=n;i++) { 65 printf("%d ",d[i]); 66 } 67 cout<<endl; 68 return 0; 69 }
十三、
树链剖分
1 //shupou 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 #define ll(x) ((x)<<1) 6 #define rr(x) ((x)<<1|1) 7 using namespace std; 8 typedef long long l; 9 10 const int N=100010;l n,m,r,p,cnt,dfscnt; 11 l d[N],id[N],f[N],son[N],top[N],siz[N],sgm[N*4],lazy[N*4],a[N],w[N],o[N],s[N*2][2]; 12 13 void jia(l x,l y) {s[++cnt][1]=o[x];s[cnt][0]=y;o[x]=cnt;} 14 15 void dfs(l x,l fa,l dep) 16 { 17 f[x]=fa;d[x]=dep;siz[x]=1;l mm=-1; 18 for (int i=o[x];i;i=s[i][1]) { 19 if (s[i][0]!=fa) { 20 dfs(s[i][0],x,dep+1);siz[x]+=siz[s[i][0]]; 21 if (siz[s[i][0]]>mm) son[x]=s[i][0],mm=siz[s[i][0]]; 22 } 23 } 24 } 25 26 void build(l x,l tp) 27 { 28 id[x]=++dfscnt;top[x]=tp;a[dfscnt]=w[x]; 29 if (son[x]) build(son[x],tp); 30 for (int i=o[x];i;i=s[i][1]) 31 if ((s[i][0]!=f[x])&&(son[x]!=s[i][0])) build(s[i][0],s[i][0]); 32 } 33 34 void xia(l r,l z,l y) 35 { 36 l k=z+y>>1;lazy[ll(r)]+=lazy[r];lazy[rr(r)]+=lazy[r]; 37 sgm[ll(r)]+=(k-z+1)*lazy[r]; 38 sgm[rr(r)]+=(y-k)*lazy[r]; 39 sgm[ll(r)]%=p;sgm[ll(r)]%=p;lazy[r]=0; 40 } 41 42 void js(l r,l z,l y) 43 { 44 if (z==y) {sgm[r]=a[z];if (sgm[r]>p) sgm[r]%=p;return;} 45 l k=z+y>>1;js(ll(r),z,k);js(rr(r),k+1,y);sgm[r]=(sgm[ll(r)]+sgm[rr(r)])%p; 46 } 47 48 void qjxg(l r,l z,l y,l zz,l yy,l v) 49 { 50 if (z>yy||y<zz) return; 51 if (z>=zz&&y<=yy) { 52 lazy[r]+=v;sgm[r]+=(y-z+1)*v;return;} 53 if (lazy[r]) xia(r,z,y);l k=z+y>>1; 54 if (z<=k) qjxg(ll(r),z,k,zz,yy,v);if (y>k) qjxg(rr(r),k+1,y,zz,yy,v); 55 sgm[r]=(sgm[ll(r)]+sgm[rr(r)])%p; 56 } 57 58 l cx(l r,l z,l y,l zz,l yy) 59 { 60 if (z>yy||y<zz) return 0; 61 if (z>=zz&&y<=yy) return sgm[r]%p; 62 l k=z+y>>1;if (lazy[r]) xia(r,z,y); 63 return (cx(ll(r),z,k,zz,yy)+cx(rr(r),k+1,y,zz,yy))%p; 64 } 65 66 l qiulu(l x,l y) 67 { 68 l ans=0;while (top[x]!=top[y]) { 69 if (d[top[x]]<d[top[y]]) swap(x,y); 70 ans+=cx(1,1,n,id[top[x]],id[x]);ans%=p;x=f[top[x]]; 71 } 72 if (d[x]>d[y]) swap(x,y);ans+=cx(1,1,n,id[x],id[y]);ans%=p; 73 return ans; 74 } 75 76 void jialu(l x,l y,l v) 77 { 78 v%=p;while (top[x]!=top[y]) { 79 if (d[top[x]]<d[top[y]]) swap(x,y); 80 qjxg(1,1,n,id[top[x]],id[x],v);x=f[top[x]]; 81 } 82 if (d[x]>d[y]) swap(x,y);qjxg(1,1,n,id[x],id[y],v); 83 } 84 85 int main() 86 { 87 l x,y,v,flag;cin>>n>>m>>r>>p;for (int i=1;i<=n;i++) scanf("%lld",&w[i]); 88 for (int i=1;i<n;i++) {scanf("%lld%lld",&x,&y);jia(x,y);jia(y,x);}dfs(r,0,1);build(r,r);js(1,1,n); 89 for (int i=1;i<=m;i++) {scanf("%lld",&flag); 90 if (flag==1) {scanf("%lld%lld%lld",&x,&y,&v);jialu(x,y,v);} 91 if (flag==2) {scanf("%lld%lld",&x,&y);printf("%lld\n",qiulu(x,y));} 92 if (flag==3) {scanf("%lld%lld",&x,&v);v%=p;qjxg(1,1,n,id[x],id[x]+siz[x]-1,v);} 93 if (flag==4) {scanf("%lld",&x);printf("%lld\n",cx(1,1,n,id[x],id[x]+siz[x]-1)%p);} 94 } 95 return 0; 96 }
十四、
最小费用最大流
1 //zxfyzdl 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 using namespace std; 6 7 const int N=5010,M=100010,inf=2e9; 8 int head,tail,cnt=-1,n,m,s,t,cost=0; 9 int q[N*10],v[N],ss[M*2][4],o[N],d[N],vis[N]; 10 11 void jia(int a,int b,int c,int cc) 12 { 13 //ss[i][0]-->to;ss[i][1]-->next;ss[i][2]-->flow;ss[i][3]-->cost; 14 ss[++cnt][0]=b;ss[cnt][1]=o[a]; 15 ss[cnt][2]=c;ss[cnt][3]=cc;o[a]=cnt; 16 ss[++cnt][0]=a;ss[cnt][1]=o[b]; 17 ss[cnt][2]=0;ss[cnt][3]=-cc;o[b]=cnt; 18 } 19 20 int dfs(int x,int flow) 21 { 22 v[x]=1;int rest=0;if (x==t) return flow; 23 for (int i=o[x];i!=-1;i=ss[i][1]) { 24 if ((d[ss[i][0]]==d[x]-ss[i][3])&&(v[ss[i][0]]==0)&&(ss[i][2]!=0)) { 25 int di=dfs(ss[i][0],min(flow-rest,ss[i][2])); 26 if (di>0) {rest+=di;ss[i][2]-=di;ss[i^1][2]+=di;} 27 } 28 } 29 return rest; 30 } 31 32 bool spfa() 33 { 34 for (int i=1;i<=n;i++) d[i]=inf,vis[i]=0; 35 head=1;tail=2;q[1]=t;d[t]=0;vis[t]=1; 36 while (head!=tail) { 37 int u=q[head]; 38 for (int i=o[u];i!=-1;i=ss[i][1]) { 39 if ((d[ss[i][0]]>d[u]-ss[i][3])&&(ss[i^1][2]!=0)) { 40 d[ss[i][0]]=d[u]-ss[i][3]; 41 if (!vis[ss[i][0]]) { 42 vis[ss[i][0]]=1;q[tail++]=ss[i][0];if (tail==n+1) tail=1; 43 } 44 } 45 } 46 q[head++]=0;vis[u]=0;if (head==n+1) head=1; 47 } 48 return d[s]!=inf; 49 } 50 51 void FU_DUI_SUAN_FA_BO_DA_JING_SHEN() 52 { 53 int ans=0,cntt=1; 54 while (spfa()) { 55 v[t]=1; 56 while (v[t]) { 57 memset(v,0,sizeof(v)); 58 int l=dfs(s,inf);ans+=l;cost+=l*d[s]; 59 } 60 } 61 cout<<ans<<" "<<cost<<endl; 62 } 63 64 int main() 65 { 66 int a,b,c,cc; 67 memset(o,-1,sizeof(o)); 68 cin>>n>>m>>s>>t; 69 for (int i=1;i<=m;i++) { 70 scanf("%d%d%d%d",&a,&b,&c,&cc); 71 jia(a,b,c,cc); 72 } 73 FU_DUI_SUAN_FA_BO_DA_JING_SHEN(); 74 return 0; 75 }
十五、
堆(手打)
1 //heap-xiao 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 using namespace std; 6 7 int n,flag,heap[1000010],heapsize; 8 9 void jiaru(int d) 10 { 11 int now,next;heap[++heapsize]=d;now=heapsize; 12 while (now>1) { 13 next=now>>1;if (heap[now]>=heap[next]) break; 14 swap(heap[now],heap[next]);now=next; 15 } 16 return; 17 } 18 19 int shanchu_quchu() 20 { 21 int now,next,res;res=heap[1];heap[1]=heap[heapsize--];now=1; 22 while (now*2<=heapsize) { 23 next=now*2; 24 if (next<heapsize&&heap[next+1]<heap[next]) next++; 25 if (heap[now]<=heap[next]) break;swap(heap[now],heap[next]); 26 now=next; 27 } 28 return res; 29 } 30 31 int main() 32 { 33 cin>>n;int x; 34 while (n--) { 35 scanf("%d",&flag); 36 if (flag==1) {scanf("%d",&x);jiaru(x);} 37 if (flag==2) {printf("%d\n",heap[1]);} 38 if (flag==3) {int daiti=shanchu_quchu();} 39 } 40 return 0; 41 }
十六、
优先队列(STL)
1 //priority_queue 2 #include <queue> 3 #include <cstdio> 4 #include <cstring> 5 #include <iostream> 6 using namespace std; 7 8 int n,flag,a; 9 priority_queue <int> heap; 10 11 int main() { 12 cin>>n; 13 for (int i=1;i<=n;i++) { 14 cin>>flag; 15 if (flag==1) { 16 cin>>a; 17 heap.push(-a); 18 } 19 else if (flag==2) { 20 int f=heap.top(); 21 cout<<-f<<endl; 22 } 23 else heap.pop(); 24 } 25 return 0; 26 }
十七、
迪杰斯特拉(堆优化版本)
1 //dijkstra-dyh 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 #define MAXN 10010 6 using namespace std; 7 typedef pair<int,int>Pair; 8 9 struct node { 10 int u,w,v,next; 11 }e[500010]; 12 13 int dis[MAXN],st[MAXN]; 14 bool flag[MAXN]; 15 int tot,start,n,m,x,y,z; 16 17 void add(int x,int y,int z) 18 { 19 e[++tot].u=x;e[tot].v=y; 20 e[tot].w=z; 21 e[tot].next=st[x];st[x]=tot; 22 } 23 24 int dijsktra(int start) 25 { 26 memset(dis,127,sizeof dis); 27 memset(flag,0,sizeof flag); 28 dis[start]=0;priority_queue< Pair,vector<Pair>,greater<Pair> >que; 29 que.push(make_pair(dis[start],start)); 30 while (!que.empty()) { 31 Pair now=que.top();que.pop(); 32 if (flag[now.second]) continue; 33 flag[now.second]=1; 34 for (int i=st[now.second];i;i=e[i].next) 35 if (dis[now.second]+e[i].w<dis[e[i].v]) { 36 dis[e[i].v]=dis[now.second]+e[i].w; 37 if (!flag[e[i].v]) que.push(make_pair(dis[e[i].v],e[i].v)); 38 } 39 } 40 for (int i=1;i<=n;i++) { 41 if (dis[i]==2139062143) dis[i]=2147483647; 42 printf("%d ",dis[i]); 43 } 44 } 45 int main() 46 { 47 scanf("%d%d%d",&n,&m,&start); 48 for (int i=1;i<=m;i++) { 49 scanf("%d%d%d",&x,&y,&z); 50 add(x,y,z); 51 } 52 dijsktra(start); 53 }
十八、
迪杰斯特拉(线段树版本)
1 //dijktra-xds 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 7 const int maxn =10007; 8 const int maxm = 500007; 9 const int INF = 0x7fffffff; 10 int n,m; 11 12 inline int read() 13 { 14 int x=0; 15 char c=getchar(); 16 while (c<‘0‘||c>‘9‘) c=getchar(); 17 while (c>=‘0‘&&c<=‘9‘) x=x*10+c-‘0‘,c=getchar(); 18 return x; 19 } 20 21 struct node{ 22 int v,next,w; 23 }edge[maxm]; 24 25 int num=0,head[maxn]; 26 27 inline void add_edge(int a,int b,int c) { 28 edge[++num].v=b;edge[num].w=c;edge[num].next=head[a];head[a]=num; 29 } 30 31 int dis[maxn],ans[maxn],s,t; 32 int tree[maxn<<2],leaf; 33 34 inline int check(int i,int j) { 35 return dis[i]<dis[j]?i:j; 36 } 37 38 inline void build() { 39 std::memset(dis,0x3f,sizeof dis); 40 for (leaf=1;leaf<=n;leaf<<=1);--leaf; 41 for (int i=1;i<=n;++i)tree[leaf+i]=i; 42 } 43 44 inline void modify(int x,int y) { 45 dis[x]=y,x+=leaf,x>>=1; 46 while(x) tree[x]=check(tree[x<<1],tree[x<<1|1]),x=x>>1; 47 } 48 49 void dijkstra(int s) { 50 build();dis[s]=0;int u=s; 51 for (int i=1;i<=n;++i) { 52 ans[u]=dis[u]; 53 const int disu=dis[u]; 54 modify(u,INF); 55 for (int j=head[u];j;j=edge[j].next){ 56 int v=edge[j].v; 57 if (dis[v]<INF&&dis[v]>disu+edge[j].w) 58 modify(v,disu+edge[j].w); 59 } 60 u=tree[1]; 61 } 62 } 63 64 inline void put(int x) 65 { 66 if (x>9) put(x/10); 67 putchar(x%10+48); 68 } 69 70 int main() { 71 int k; 72 n=read(),m=read(),k=read(); 73 for (int a,b,c,i=1;i<=m;++i) { 74 a=read(),b=read(),c=read(); 75 add_edge(a,b,c); 76 } 77 dijkstra(k); 78 for (int i=1;i<=n;++i) { 79 if (dis[i]==0x3f3f3f3f) ans[i]=INF; 80 put(ans[i]);putchar(‘ ‘); 81 } 82 putchar(‘\n‘); 83 return 0; 84 }
十九、普通平衡树(splay)
1 //pinghengshu 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 using namespace std; 6 typedef long long ll; 7 8 const int N=100010; 9 int ch[N][2],f[N],ky[N],ct[N],siz[N],sz,rt; 10 11 void clear(int x) 12 { 13 ch[x][0]=ch[x][1]=f[x]=siz[x]=ct[x]=ky[x]=0; 14 return; 15 } 16 17 bool get(int x) 18 { 19 return ch[f[x]][1]==x; 20 } 21 22 void update(int x) 23 { 24 if (x) { 25 siz[x]=ct[x]; 26 if (ch[x][0]) siz[x]+=siz[ch[x][0]]; 27 if (ch[x][1]) siz[x]+=siz[ch[x][1]]; 28 } 29 return; 30 } 31 32 void rotate(int x) 33 { 34 int old=f[x],oldf=f[old],which=get(x); 35 ch[old][which]=ch[x][which^1];f[ch[old][which]]=old; 36 ch[x][which^1]=old;f[old]=x;f[x]=oldf; 37 if (oldf) ch[oldf][ch[oldf][1]==old]=x; 38 update(old);update(x); 39 } 40 41 void splay(int x) 42 { 43 for (int fa;fa=f[x];rotate(x)) 44 if (f[fa]) rotate((get(x)==get(fa))?fa:x); 45 rt=x;return; 46 } 47 48 void insert(int v) 49 { 50 if (!rt) { 51 sz++;ch[sz][0]=ch[sz][1]=f[sz]=0; 52 ky[sz]=v;ct[sz]=1;siz[sz]=1;rt=sz; 53 return; 54 } 55 int now=rt,fa=0; 56 while (1) { 57 if (ky[now]==v) { 58 ct[now]++; 59 update(now);update(fa); 60 splay(now);break; 61 } 62 fa=now;now=ch[now][ky[now]<v]; 63 if (!now) { 64 sz++;ch[sz][0]=ch[sz][1]=0;ky[sz]=v;siz[sz]=1; 65 ct[sz]=1;f[sz]=fa;ch[fa][ky[fa]<v]=sz; 66 update(fa);splay(sz);break; 67 } 68 } 69 return; 70 } 71 72 int find(int v) 73 { 74 int ans=0,now=rt; 75 while (1) { 76 if (v<ky[now]) now=ch[now][0]; 77 else { 78 ans+=(ch[now][0]?siz[ch[now][0]]:0); 79 if (v==ky[now]) {splay(now);return ans+1;} 80 ans+=ct[now];now=ch[now][1]; 81 } 82 } 83 return 0; 84 } 85 86 int findx(int x) 87 { 88 int now=rt; 89 while (1) { 90 if (ch[now][0]&&x<=siz[ch[now][0]]) now=ch[now][0]; 91 else { 92 int tt=(ch[now][0]?siz[ch[now][0]]:0)+ct[now]; 93 if (x<=tt) return ky[now]; 94 x-=tt;now=ch[now][1]; 95 } 96 } 97 return 0; 98 } 99 100 int pre() 101 { 102 int now=ch[rt][0]; 103 while (ch[now][1]) now=ch[now][1]; 104 return now; 105 } 106 107 int next() 108 { 109 int now=ch[rt][1]; 110 while (ch[now][0]) now=ch[now][0]; 111 return now; 112 } 113 114 void del(int x) 115 { 116 int whatever=find(x); 117 if (ct[rt]>1) {ct[rt]--;update(rt);return;} 118 if (!ch[rt][0]&&!ch[rt][1]) {clear(rt);rt=0;return;} 119 if (!ch[rt][0]) { 120 int oldrt=rt;rt=ch[rt][1];f[rt]=0;clear(oldrt);return; 121 } 122 else if (!ch[rt][1]) { 123 int oldrt=rt;rt=ch[rt][0];f[rt]=0;clear(oldrt);return; 124 } 125 int leftbig=pre(),oldrt=rt; 126 splay(leftbig); 127 ch[rt][1]=ch[oldrt][1];f[ch[oldrt][1]]=rt; 128 clear(oldrt);update(rt);return; 129 } 130 131 int main() 132 { 133 int n,a,b;cin>>n;int rrr=0; 134 while (n--) { 135 scanf("%d%d",&a,&b); 136 if (a==1) insert(b)/*,cout<<++rrr<<"i"<<endl*/; 137 else if (a==2) del(b)/*,cout<<++rrr<<"d"<<endl*/; 138 else if (a==3) printf("%d\n",find(b))/*,cout<<++rrr<<"f"<<endl*/; 139 else if (a==4) printf("%d\n",findx(b))/*,cout<<++rrr<<"x"<<endl*/; 140 else if (a==5) {insert(b);printf("%d\n",ky[pre()]);del(b)/*,cout<<++rrr<<"p"<<endl*/;} 141 else {insert(b);printf("%d\n",ky[next()]);del(b)/*,cout<<++rrr<<"n"<<endl*/;} 142 } 143 return 0; 144 }
二十、矩阵快速幂
1 //jzksm 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 using namespace std; 6 7 long long n,p; 8 long long mod=1e9+7; 9 long long ans[105][105]; 10 long long a[105][105],b[105][105]; 11 12 void lu() { 13 memcpy(b,ans,sizeof(ans)); 14 memset(ans,0,sizeof(ans)); 15 for (int i=1;i<=n;i++) 16 for (int j=1;j<=n;j++) 17 for (int k=1;k<=n;k++) 18 ans[i][j]=(ans[i][j]+(b[i][k]*a[k][j])%mod)%mod; 19 } 20 21 void ge() { 22 memcpy(b,a,sizeof(a)); 23 memset(a,0,sizeof(a)); 24 for (int i=1;i<=n;i++) 25 for (int j=1;j<=n;j++) 26 for (int k=1;k<=n;k++) 27 a[i][j]=(a[i][j]+(b[i][k]*b[k][j])%mod)%mod; 28 } 29 30 int main() { 31 cin>>n>>p;p--; 32 for (int i=1;i<=n;i++) 33 for (int j=1;j<=n;j++) 34 cin>>ans[i][j]; 35 memcpy(a,ans,sizeof(a)); 36 while (p) { 37 if (p&1) lu(); 38 ge();p>>=1; 39 } 40 for (int i=1;i<=n;i++) { 41 for (int j=1;j<=n;j++) 42 cout<<ans[i][j]<<‘ ‘; 43 cout<<endl; 44 } 45 return 0; 46 }
未完待续。。。