pat advanced 1013
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地址:http://pat.zju.edu.cn/contests/pat-a-practise/1013
思路:通过dfs来判断图的联通性,通过N次dfs才遍历完图所有节点的话,那么就有N-1个不连同图。
参考代码:
#include<cstdio>#include<cstring>int N, M, lost;bool map[1000][1000], visited[1000];void dfs(int start){ for(int i = 1; i <= N; ++i) { if(i != lost && i != start && !visited[i] && map[start][i]) { visited[i] = true; dfs(i); } }}int main(){ int K, c1, c2, cnt; scanf("%d %d %d", &N, &M, &K); for(int i = 0; i < M; ++i) { scanf("%d %d", &c1, &c2); map[c1][c2] = map[c2][c1] = true; } for(int i = 0; i < K; ++i) { scanf("%d", &lost); cnt = 0; memset(visited, 0, sizeof(visited)); for(int j = 1; j<=N; ++j) { if(j != lost && visited[j]==false) { ++cnt; dfs(j); } } printf("%d\n", cnt-1); } return 0;}
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