2018 Spring Single Training B （uva 572，HihoCoder 1632，POJ 2387，POJ 2236，UVA 10054，HDU 2141）

Posted 2020-10-30 蒟蒻LQL

这场比赛可以说是灰常的水了，涨信心场？？

今下午义务劳动，去拿着锄头发了将近一小时呆，发现自己实在是干不了什么，就跑到实验室打比赛了~

之前的比赛补题补了这么久连一场完整的都没补完，结果这场比完后一小时连题解都出来了···

A-烤肉拌饭 （ uva 572）

就是求联通块的数量啊，刚学dfs的时候做的那种！

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <queue>
 6 
 7 using namespace std;
 8 const int maxn=100+5;
 9 const int dx[]={1,-1,1,-1,1,-1,0,0};
10 const int dy[]={0,0,1,1,-1,-1,1,-1};
11 char G[maxn][maxn];
12 int vis[maxn][maxn];
13 int n,m;
14 int ans;
15 void dfs(int x,int y){
16     vis[x][y]=1;
17     for(int i=0;i<8;i++){
18         int nx=x+dx[i];
19         int ny=y+dy[i];
20         if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&G[nx][ny]==‘@‘&&!vis[nx][ny]){
21             dfs(nx,ny);
22         }
23     }
24     return ;
25 }
26 int main(){
27     while(scanf("%d%d",&n,&m)&&n&&m){
28         memset(vis,0,sizeof(vis));
29         ans=0;
30         for(int i=1;i<=n;i++){
31             for(int j=1;j<=m;j++){
32                 scanf(" %c",&G[i][j]);
33             }
34         }
35         for(int i=1;i<=n;i++){
36             for(int j=1;j<=m;j++){
37                 if(!vis[i][j]&&G[i][j]==‘@‘){
38                     ans++;
39                     dfs(i,j);
40                 }
41             }
42         }
43         cout<<ans<<endl;
44     }
45 return 0;
46 }

View Code

 

 

B-麻辣香锅 HihoCoder - 1632

模拟，emmm模拟可以提高代码能力？？

C-啵啵鱼 poj2387

是个很裸的最短路，一个spfa或者dijsktra就可以

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <queue>
 6 
 7 using namespace std;
 8 const int maxn=2000+10;
 9 const int maxm=4000+10;
10 const int INF=2147483600;
11 typedef long long LL;
12 int head[maxn],Next[maxm],to[maxm];
13 LL val[maxm];
14 int sz;
15 int n,m;
16 void add_edge(int a,int b,LL w){
17     ++sz;
18     val[sz]=w;
19     to[sz]=b;
20     Next[sz]=head[a];
21     head[a]=sz;
22 }
23 LL d[maxn];
24 int spfa(int s){
25     for(int i=1;i<=n;i++)d[i]=INF;
26     int vis[maxn];
27     memset(vis,0,sizeof(vis));
28     queue<int>q;
29     q.push(s);
30     vis[s]=1;
31     d[s]=0;
32     while(!q.empty()){
33         int u=q.front();q.pop();vis[u]=0;
34         for(int i=head[u];i;i=Next[i]){
35             int v=to[i];
36             if(d[v]>d[u]+val[i]){
37                 d[v]=d[u]+val[i];
38                 if(!vis[v]){
39                     vis[v]=1;
40                     q.push(v);
41                 }
42             }
43         }
44     }
45     return d[1];
46 }
47 int main(){
48     sz=0;
49     memset(head,0,sizeof(head));
50     scanf("%d%d",&m,&n);
51     int a,b;LL c;
52     for(int i=1;i<=m;i++){
53         scanf("%d %d%lld",&a,&b,&c);
54       //  cout<<a<<" "<<b<<" "<<c<<endl;
55         add_edge(a,b,c);
56         add_edge(b,a,c);
57     }
58     cout<<spfa(n);
59 return 0;
60 }

View Code

D-海底捞 POJ2236

卢老师一句话点醒梦中人，直接并查集就可以~

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <cstring>
 5 using namespace std;
 6 const int maxn=1000+10;
 7 int n,d;
 8 int a,b;
 9 char c;
10 int p[maxn],vis[maxn];
11 struct Node{
12     int x,y;
13 }node[maxn];
14 int find(int x){
15     return p[x]==x?x:p[x]=find(p[x]);
16 }
17 bool dist(Node A,Node B){
18     if((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y)<=d*d)
19         return true;
20     return false;
21 }
22 int main(){
23     scanf("%d%d",&n,&d);
24     for(int i=1;i<=n;i++){
25         scanf("%d%d",&node[i].x,&node[i].y);
26     }
27     for(int i=0;i<=n;i++)p[i]=i;
28     while(scanf(" %c",&c)!=EOF){
29         if(c==‘O‘){
30             scanf("%d",&a);
31             vis[a]=1;
32             for(int i=1;i<=n;i++){
33                 if(vis[i]&&i!=a&&dist(node[a],node[i])){
34                    // cout<<a<<"-->"<<i<<endl;
35                     int x=find(a),y=find(i);
36                     if(x!=y)p[x]=y;
37                 }
38             }
39         }
40         if(c==‘S‘){
41             scanf("%d%d",&a,&b);
42             int x=find(a),y=find(b);
43             //cout<<":"<<x<<y<<endl;
44             if(x==y){
45                 printf("SUCCESS\n");
46             }else{
47                 printf("FAIL\n");
48             }
49         }
50     }
51     return 0;
52 }

View Code

E-黄焖鸡 uva10054

这个题以前做过，欧拉回路，注意打印的时候逆序输出

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <queue>
 6 
 7 using namespace std;
 8 const int maxn=1000+20;
 9 
10 int T;
11 int G[maxn][maxn];
12 int n;
13 int deg[maxn];
14 int a,b;
15 int M;
16 int p[maxn];
17 int find(int x){
18     return p[x]==x?x:find(p[x]);
19 }
20 void dfs(int u){
21     for(int i=1;i<=M;i++){
22         if(G[u][i]){
23             G[u][i]--;
24             G[i][u]--;
25             dfs(i);
26             cout<<i<<" "<<u<<endl;
27         }
28     }
29     return;
30 }
31 int main(){
32     scanf("%d",&T);
33     for(int t=1;t<=T;t++){
34         if(t!=1)cout<<""<<endl;
35         memset(G,0,sizeof(G));
36         memset(deg,0,sizeof(deg));
37         printf("Case #%d\n",t);
38         M=-1;
39         scanf("%d",&n);
40         for(int i=1;i<=55;i++)p[i]=i;
41         for(int i=1;i<=n;i++){
42             scanf("%d%d",&a,&b);
43             M=max(M,max(a,b));
44             G[a][b]++;
45             G[b][a]++;
46             deg[a]++;
47             deg[b]++;
48             int x=find(a),y=find(b);
49             if(x!=y)p[x]=y;
50 
51         }
52         bool judge=1;
53         for(int i=1;i<=M;i++){
54             if(deg[i]){
55                 if(find(i)!=find(M)||deg[i]%2){
56                     judge=0;
57                     break;
58                 }
59             }
60         }
61         if(!judge){
62             cout<<"some beads may be lost"<<endl;
63         }else{
64         for(int i=1;i<=M;i++)
65             dfs(i);
66         }
67 
68     }
69 return 0;
70 }

View Code

F- 减肥成功指日可待 hdu2141

用set或者二分。但是set再G++下会MLE必须要选择C++才能AC，雾···

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <set>
 6 #include <vector>
 7 using namespace std;
 8 const int maxn=500+10;
 9 typedef long long LL;
10 int A,B,C,s,x;
11 LL a[maxn],b[maxn],c[maxn];
12 int kase;
13 int main(){
14     while(scanf("%d%d%d",&A,&B,&C)!=EOF){
15     kase++;
16     set<LL>S;
17     printf("Case %d:\n",kase);
18     for(int i=1;i<=A;i++){
19         scanf("%lld",&a[i]);
20     }
21     for(int i=1;i<=B;i++){
22         scanf("%lld",&b[i]);
23     }
24     for(int i=1;i<=C;i++){
25         scanf("%lld",&c[i]);
26     }
27     for(int i=1;i<=A;i++){
28         for(int j=1;j<=B;j++){
29             //for(int l=1;l<=C;l++){
30                 LL X=a[i]+b[j];
31                 if(!S.count(X))
32                     S.insert(X);
33            // }
34         }
35     }
36     scanf("%d",&s);
37     LL x;
38     for(int i=1;i<=s;i++){
39         scanf("%lld",&x);
40         bool ok=0;
41         for(int i=1;i<=C;i++){
42             LL pos=x-c[i];
43             //if(pos<0)continue;
44             if(S.count(pos)){
45                 ok=1;
46                 break;
47             }
48         }
49         //if(S.count(x))cout<<"YES"<<endl;
50         if(ok)cout<<"YES"<<endl;
51         else
52             cout<<"NO"<<endl;
53     }
54     }
55 return 0;
56 }

View Code