hdu 5438 Ponds（并查集+拓扑序）

来源：互联网发布：淘宝人像摄影教程编辑：程序博客网时间：2024/04/28 01:17

题目链接：hdu 5438 Ponds

先把点按照类似拓扑排序的方法删掉，然后对剩下的点用并查集维护，并且维护联通分量的点个数和权值和。

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 1e4 + 5;int N, M, L[maxn], R[maxn], F[maxn], C[maxn], in[maxn], vi[maxn];ll W[maxn];vector<int> G[maxn];int find (int x) { return x == F[x] ? x : F[x] = find(F[x]); }void init () {scanf("%d%d", &N, &M);memset(in, 0, sizeof(in));memset(vi, 0, sizeof(vi));for (int i = 1; i <= N; i++) {G[i].clear();scanf("%lld", &W[i]);}int u, v;for (int i = 0; i < M; i++) {scanf("%d%d", &u, &v);G[u].push_back(v);G[v].push_back(u);in[v]++, in[u]++;L[i] = u; R[i] = v;}}ll solve () {queue<int> Q;for (int i = 1; i <= N; i++) if (in[i] < 2) {vi[i] = 1;Q.push(i);}while (!Q.empty()) {int u = Q.front();Q.pop();for (int i = 0; i < G[u].size(); i++) {int v = G[u][i];if (vi[v]) continue;in[v]--;if (in[v] < 2) {vi[v] = 1;Q.push(v);}}}for (int i = 0; i <= N; i++)F[i] = i, C[i] = 1;for (int i = 0; i < M; i++) {int u = L[i], v = R[i];if (vi[u] || vi[v]) continue;if (find(u) != find(v)) {C[find(v)] += C[find(u)];W[find(v)] += W[find(u)];F[find(u)] = find(v);}}ll ans = 0;for (int i = 1; i <= N; i++) {if (vi[i] || F[i] != i) continue;if (C[i]&1)ans += W[i];}return ans;}int main () {int cas;scanf("%d", &cas);while (cas--) {init();printf("%lld\n", solve());}return 0;}

0 0