2017-10-20 NOIP模拟赛
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Lucky Transformation
#include<iostream> #include<cstring> #include<cstdio> using namespace std; int n,k,cnt; char s[1000010]; int main(){ freopen("trans.in","r",stdin);freopen("trans.out","w",stdout); //freopen("Cola.txt","r",stdin); while(scanf("%d%d",&n,&k)!=EOF){ scanf("%s",s);cnt=0; int rest=0; for(int i=0;i<n;i++){ if(s[i]==\'2\'&&s[i+1]==\'3\'){ if(i>0&&(i+1)%2==0&&s[i-1]==\'2\'&&s[i+1]==\'3\'){ rest=k-cnt;break; } else if((i+1)%2==1&&s[i+1]==\'3\'&&s[i+2]==\'3\'){ rest=k-cnt;break; } else{ if((i+1)%2==0)s[i]=\'3\'; else s[i+1]=\'2\'; cnt++; if(cnt==k)break; } } } rest%=2; if(rest==0||cnt==k){ printf("%s\\n",s); } else{ for(int i=0;i<n;i++){ if(s[i]==\'2\'&&s[i+1]==\'3\'){ if((i+1)%2==0)s[i]=\'3\'; else s[i+1]=\'2\'; break; } } printf("%s\\n",s); } } }
Snake vs Block
#include<iostream> #include<cstdio> #define maxn 210 using namespace std; int n,m,map[maxn][7],num,ans,head[maxn],edge[maxn*6+10][maxn*6+10]; struct node{ int to,pre,v; }e[maxn*7*3]; int make_id(int i,int j){return (i-1)*5+j;} void Insert(int from,int to,int v){ e[++num].to=to; e[num].v=v; e[num].pre=head[from]; head[from]=num; edge[from][to]=num; } void dfs(int x,int y,int sc,int len){ if(x>n){ ans=max(ans,sc); return; } int now=make_id(x,y); for(int i=head[now];i;i=e[i].pre){ int to=e[i].to; if(e[i].v==0)continue; int xx=(to-1)/5+1,yy=to%5; if(yy==0)yy=5; e[i].v=0; if(map[xx][yy]>=0){ int p=map[xx][yy]; map[xx][yy]=0; dfs(xx,yy,sc,len+p); map[xx][yy]=p; } else { if(len+map[xx][yy]<0){ans=max(ans,sc);} else { int p=map[xx][yy]; map[xx][yy]=0; dfs(xx,yy,sc-p,len+p); map[xx][yy]=p; } } e[i].v=1; } } int main(){ freopen("snakevsblock.in","r",stdin);freopen("snakevsblock.out","w",stdout); //freopen("Cola.txt","r",stdin); scanf("%d",&n); for(int i=1;i<=n;i++) for(int j=1;j<=5;j++){ int id=make_id(i,j); if(j!=1)Insert(id,make_id(i,j-1),1);//向左 if(j!=5)Insert(id,make_id(i,j+1),1);//向右 Insert(id,make_id(i+1,j),1);//向下 scanf("%d",&map[i][j]); } scanf("%d",&m); int x,y; for(int i=1;i<=m;i++){ scanf("%d%d",&x,&y); int id1=make_id(x,y),id2=make_id(x,y+1); e[edge[id1][id2]].v=0; e[edge[id2][id1]].v=0; } dfs(1,3,0,4); printf("%d",ans); }
/* 把豆豆和砖块丢在一起dp f[i][j][k]:到第i行,蛇的长度还剩下j,从k位置出第i行的最大得分 g[j][l][r]:蛇的长度剩下j,从当前行l到r区间内仍为死亡的最大得分 初始化:对于每个i，g[j][k][k]=f[i-1][j-a[i][k]][k]+max(-a[i][k],0) 为初始状态 方程g[j][l][r] = max(g[j-a[i][l]][l+1][r] + max(-a[i][j],0),g[j-a[i][r]][l][r-1]+max(-a[i][k],0)); 最后f[i][j][k] = max{g[j][l][r](1≤l≤k ≤r ≤5)} */ #include<iostream> #include<cstdio> #include<cstring> #define N 207 #define M 10005 using namespace std; int a[N][5],f[N][M][5],g[M][5][5]; bool flag[N][4]; int n,m,maxi,ans,val; inline int read() { int x=0,f=1;char c=getchar(); while(c>\'9\'||c<\'0\'){if(c==\'-\')f=-1;c=getchar();} while(c>=\'0\'&&c<=\'9\'){x=x*10+c-\'0\';c=getchar();} return x*f; } int main() { freopen("snakevsblock.in","r",stdin); freopen("snakevsblock.out","w",stdout); n=read();int x,y; for(int i=1;i<=n;i++) for(int j=0;j<5;j++) a[i][j]=read(); m=read(); for(int i=1;i<=m;i++) { x=read();y=read(); flag[x][y-1]=1; } memset(f,-0x7f7f7f,sizeof f); f[0][4][2]=0;maxi=n*50; for(int i=1;i<=n;i++) { memset(g,-0x7f7f7f,sizeof g); for(int j=0;j<=maxi;j++) for(int k=0;k<5;k++) if(j-a[i][k]>=0 && j-a[i][k]<=maxi) f[i][j][k]=g[j][k][k]=f[i-1][j-a[i][k]][k]+max(-a[i][k],0); for(int l=1;l<=4;l++) for(int j=0,k=j+l;k<5;j++,k++) for(int v=0;v<=maxi;v++) { if(!flag[i][j] && (val=v-a[i][j])>=0 && val<=maxi) g[v][j][k] = g[val][j + 1][k] + max(-a[i][j], 0);//g[v][j][k]=max(g[v][j][k],g[val][j+1][k]+max(-a[i][j],0)); if(!flag[i][k-1]&& (val=v-a[i][k])>=0 && val<=maxi) g[v][j][k]=max(g[v][j][k],g[val][j][k-1]+max(-a[i][k],0)); for(int to=j;to<=k;to++) f[i][v][to]=max(f[i][v][to],g[v][j][k]); } } for(int i=0;i<=n;i++) for(int j=0;j<=maxi;j++) for(int k=0;k<5;k++) ans=max(f[i][j][k],ans); printf("%d\\n",ans); return 0; }
Ping
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#include<iostream> #include<cstring> #include<cstdio> #define maxn 100010 using namespace std; int n,m,k,num,head[maxn],fa[maxn],dep[maxn],h[maxn],cnt; bool vis[maxn],v[maxn]; int c[16][1<<15]; struct node{ int to,pre; }e[maxn*2]; struct Node{ int l,r; }p[300010]; void Insert(int from,int to){ e[++num].to=to; e[num].pre=head[from]; head[from]=num; } int count(int x){ int res=0; while(x){ if(x&1)res++; x>>=1; } return res; } void dfs(int now,int father){ fa[now]=father;dep[now]=dep[father]+1; for(int i=head[now];i;i=e[i].pre){ int to=e[i].to; if(to==father)continue; dfs(to,now); } } int sta; bool lca(int a,int b){ if(vis[a]||vis[b])return 1; if(dep[a]<dep[b])swap(a,b); while(dep[a]>dep[b]){ a=fa[a]; if(vis[a])return 1; } while(a!=b){ a=fa[a];b=fa[b]; if(vis[a]||vis[b])return 1; } return 0; } bool ok(int s){ memset(vis,0,sizeof(vis)); int pos=0; while(s){ pos++; if(s&1)vis[h[pos]]=1; s>>=1; } for(int i=1;i<=k;i++){ int a=p[i].l,b=p[i].r; if(lca(a,b))continue; else return 0; } return 1; } bool check(int x){ for(int i=1;i<=c[x][0];i++){ int s=c[x][i]; if(ok(s)){ sta=s; return 1; } } return 0; } int main(){ freopen("ping.in","r",stdin);freopen("ping.out","w",stdout); //freopen("Cola.txt","r",stdin); scanf("%d%d",&n,&m); for(int i=0;i<=(1<<n)-1;i++){ int nu=count(i); c[nu][++c[nu][0]]=i; } int x,y; for(int i=1;i<=m;i++){ scanf("%d%d",&x,&y); Insert(x,y);Insert(y,x); if(!v[x])h[++cnt]=x,v[x]=1; if(!v[y])h[++cnt]=y,v[y]=1; } dfs(1,1); scanf("%d",&k); for(int i=1;i<=k;i++) scanf("%d%d",&p[i].l,&p[i].r); int l=1,r=cnt,ans=0; while(l<=r){ int mid=(l+r)>>1; if(check(mid))ans=mid,r=mid-1; else l=mid+1; } printf("%d\\n",ans); int pos=0; while(sta){ pos++; if(sta&1)printf("%d ",h[pos]); sta>>=1; } return 0; }
#include <bits/stdc++.h> using namespace std; int n, m, P, u, v, to[200005], nxt[200005], p[100005], deep[100005], q[500005][2]; int son[100005], fa[100005], size[100005], top[100005], dfsx[100005], cnt, ans; int lca[500005], Q[500005], lr[100005][2]; int sta[100005]; bool flag[400005]; void dfs(int x) { size[x] = 1; for (int i = p[x]; i != -1; i = nxt[i]) if (to[i] != fa[x]) { fa[to[i]] = x; deep[to[i]] = deep[x] + 1; dfs(to[i]); if (son[x] == -1 || size[to[i]] > size[son[x]]) son[x] = to[i]; size[x] += size[to[i]]; } } void dfs1(int x) { dfsx[x] = ++cnt; if (son[x] != -1) top[son[x]] = top[x], dfs1(son[x]); for (int i = p[x]; i != -1; i = nxt[i]) if (to[i] != fa[x] && to[i] != son[x]) top[to[i]] = to[i], dfs1(to[i]); } int findlca(int x, int y) { while (1) { if (top[x] == top[y]) return deep[x] > deep[y]? y : x; if (deep[top[x]] > deep[top[y]]) x = fa[top[x]]; else y = fa[top[y]]; } } bool query(int x, int l, int r, int ll, int rr) { if (l == ll && r == rr) return flag[x]; int mid = (l + r) >> 1, L = x << 1, R = L | 1; if (rr <= mid) return query(L, l, mid, ll, rr); else if (ll > mid) return query(R, mid + 1, r, ll, rr); else return query(L, l, mid, ll, mid) | query(R, mid + 1, r, mid + 1, rr); } void modify(int x, int l, int r, int to) { flag[x] = true; if (l == r) return; int mid = (l + r) >> 1, L = x << 1, R = L | 1; if (to <= mid) modify(L, l, mid, to); else modify(R, mid + 1, r, to); } bool Query(int x, int y) { while (1) { if (top[x] == top[y]) { if (deep[x] < deep[y]) return query(1, 1, cnt, dfsx[x], dfsx[y]); else return query(1, 1, cnt, dfsx[y], dfsx[x]); } if (deep[top[x]] > deep[top[y]]) if (query(1, 1, cnt, dfsx[top[x]], dfsx[x])) return true; else x = fa[top[x]]; else { if (query(1, 1, cnt, dfsx[top[y]], dfsx[y])) return true; else y = fa[top[y]]; } } } void work(int x) { for (int i = p[x]; i != -1; i = nxt[i]) if (to[i] != fa[x]) work(to[i]); for (int i = lr[x][0]; i <= lr[x][1]; i++) if (!Query(q[Q[i]][0], q[Q[i]][1])) { modify(1, 1, cnt, dfsx[x]); sta[++ans] = x; return; } } bool cmp(int x, int y) {return lca[x] < lca[y];} int main() { freopen("ping.in","r",stdin); freopen("ping.out","w",stdout); scanf("%d%d", &n, &m); for (int i = 0; i <= n; i++) p[i] = son[i] = -1, top[i] = size[i] = fa[i] = deep[i] = 0; for (int i = 1; i <= n * 4; i++) flag[i] = false; for (int i = 1; i <= m; i++) { scanf("%d%d", &u, &v); u--, v--; to[i * 2 - 1] = v; nxt[i * 2 - 1] = p[u]; p[u] = i * 2 - 1; to[i * 2] = u; nxt[i * 2] = p[v]; p[v] = i * 2; } deep[0] = 1; dfs(0); cnt = 0; dfs1(0); scanf("%d", &P); for (int i = 1; i <= P; i++) { scanf("%d%d", &u, &v); u--, v--; q[i][0] = u, q[i][1] = v; lca[i] = findlca(u, v); Q[i] = i; } sort(Q + 1, Q + P + 1, cmp); for (int i = 0; i <= n; i++) lr[i][0] = P + 1, lr[i][1] = 0; for (int i = 1; i <= P; i++) { lr[lca[Q[i]]][0] = min(lr[lca[Q[i]]][0], i); lr[lca[Q[i]]][1] = max(lr[lca[Q[i]]][1], i); } ans = 0; work(0); printf("%d\\n", ans); for (int i = 1; i <= ans; i++) printf("%d ", sta以上是关于2017-10-20 NOIP模拟赛的主要内容，如果未能解决你的问题，请参考以下文章