题意：给了你一张图，问是否 对任意的A，B，C，有A-B 和A-C的话，同时有B-C
分析：其实就是每个块都是一个完全图，n个顶点的完全图有n*(n-1)/2条边。用并查集统计每个块有几个点，几条边。
需要注意的是n*(n-1)/2这里是会爆int的，需要用longlong(心痛。。
以下是代码：
#include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define ull unsigned long long #define ll long long #define lson l,mid,id<<1 #define rson mid+1,r,id<<1|1 typedef pairpii; typedef pairpll; typedef pairpdd; const double eps = 1e-6; const int MAXN = 1005; const int MAXM = 5005; const ll LINF = 0x3f3f3f3f3f3f3f3f; const int INF = 0x3f3f3f3f; const int MOD = 1000000007; const double FINF = 1e18; int pre[150005]; int find(int x) { return x == pre[x] ? x : pre[x] = find(pre[x]); } void mix(int a, int b) { int fa = find(a), fb = find(b); if (fa != fb) pre[fa] = fb; } int x[150005], y[150005]; int cnt[150005]; int pp[150005]; int main() { ll n, m; memset(cnt, 0, sizeof(cnt)); memset(pp, 0, sizeof(pp)); cin >> n >> m; for (int i = 1; i <= n; ++i)pre[i] = i; for (int i = 0; i < m; ++i) { scanf("%d%d", &x[i], &y[i]); mix(x[i], y[i]); } for (int i = 0; i < m; ++i) { cnt[find(x[i])]++; } for (int i = 1; i <= n; ++i) { pp[find(i)]++; } for (int i = 1; i <= n; ++i) { if (find(i) == i) { if (pp[i] *1LL* (pp[i] - 1) / 2 != cnt[i]) { printf("NO\n"); return 0; } } } printf("YES\n"); }
