hdu----King's Phone

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题意：给你n个点，有n+1条边，现在要删除一些边（至少一条），最后这些点还是可以都联通。

分析：要使n个点都联通，至少需要n-1条边，所以至多删除2条边。

枚举这两条边（或者唯一的一条边）然后用并查集判断所有的点是否相连。

复杂度O（N^3)；

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>using namespace std;int Father[111];int from[111];int to[111];int Find(int x){return Father[x] == x ? x : Father[x] = Find(Father[x]);}void Union(int x, int y){int xx = Find(x);int yy = Find(y);if (xx != yy){Father[xx] = yy;}}bool Judge(int n,int m,int x,int y){for (int i = 1; i <= n; i++) Father[i] = i;    //初始化 for (int i = 1; i <= m; i++){if (i == x || i == y) continue;           //这条边已经删掉了Union(from[i], to[i]);}//判断所有的点是否相连for (int i = 1; i <= n; i++){int cur = Find(i);if (cur != Find(1)){return false;}}return true;}int main(){int TestCas;scanf("%d", &TestCas);while (TestCas--){int n, m;scanf("%d", &n);m = n + 1;for (int i = 1; i <= m; i++){scanf("%d%d", &from[i], &to[i]);}//删除一条边int ans = 0;for (int i = 1; i <= m; i++){if (Judge(n, m, i, -1) == true)ans++;}for (int i = 1; i <= m; i++){for (int j =i + 1; j <= m; j++){if (Judge(n, m, i, j) == true){ans++;}}}printf("%d\n", ans);}return 0;}

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