UVa 10985 Rings'n'Ropes floyd+BFS

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n个戒指和m条长度为1 的绳子

求左右手拿住任意2个戒指最多几条绳子被拉直

拿住哪2个不知道所以两两枚举

对于2个戒指之间可以考虑最短路然后可能有很多条最短路在路径统计下就行

看懂样例怎么出来的就知道怎么回事了

样例3就是一个环拉住0 和 2 你们就有2条最短路每条都是3

样例4是一样的

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int MAX = 130;int a[MAX][MAX];int map[MAX][MAX];bool vis[MAX][MAX];int n,m;int cnt;void floyd(){int i,j,k;for(k = 0; k < n; k++)for(i = 0; i < n; i++)for(j = 0; j < n; j++)a[i][j] = min(a[i][j],a[i][k]+a[k][j]);}void bfs(int s,int e){queue <int> q;q.push(s);int i;while(!q.empty()){int u = q.front();q.pop();for(i = 0; i < n; i++){if(vis[u][i])continue;if(!map[u][i])continue;if(a[e][i] + 1 == a[e][u]){cnt++;vis[u][i] = vis[i][u] = true;q.push(i);}}}}int main(){int t,cas = 1;int i,j;int s,e;scanf("%d",&t);while(t--){memset(a,0,sizeof(a));memset(map,0,sizeof(map));scanf("%d %d",&n,&m);for(i = 0; i < n; i++)for(j = 0; j < n; j++){if(i == j)a[i][j] = 0;elsea[i][j] = 999999999;}for(i = 0; i < m; i++){scanf("%d %d",&s,&e);a[s][e] = a[e][s] = 1;map[s][e] = map[e][s] = 1;}floyd();int max = 0;for(i = 0; i < n; i++){for(j = i+1; j < n; j++){if(a[i][j] == 999999999)continue;memset(vis,false,sizeof(vis));cnt = 0;bfs(i,j);if(max < cnt)max = cnt;}}//printf("%d %d\n",a[0][1],a[1][0]);printf("Case #%d: %d\n",cas++,max);}return 0;}

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