題意：

給出一個無向圖和一個此圖的生成樹，讓我們求一個本圖的最小割邊集，割邊集只包括生成樹中的一條邊，讓割邊集的邊數最少，輸出數目。

思路：

并查集+暴力。

代碼：

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include <vector> using namespace std;const int maxn = 20005; const int maxm = 400005;struct Node {int to, next; }e1[maxm], e2[maxm]; int head1[maxn], head2[maxn]; int tot1, tot2;void add1(int u, int v) {e1[tot1].to = v;e1[tot1].next = head1[u]; head1[u] = tot1++; }void add2(int u, int v) {e2[tot2].to = v;e2[tot2].next = head2[u]; head2[u] = tot2++; }int n;int vis1[maxn]; int num[maxn]; int ans; int sum; int vis2[maxn];vector<int> vv[maxn];int fa[maxn]; int find(int x) {if(x == fa[x]) return fa[x];return fa[x] = find(fa[x]); } void unin(int x, int y) {int fx = find(x), fy = find(y);if(fx != fy) {fa[fy] = fx;} }void dfs(int u) {vis1[u] = 1;bool flag = false;for(int i = head1[u]; i; i = e1[i].next) {int v = e1[i].to;if(!vis1[v]) {dfs(v);unin(u, v);for(int j = 0; j < vv[v].size(); j++) {if(find(u) != find(vv[v][j])) {vv[u].push_back(vv[v][j]);}}flag = true;}}for(int i = head2[u]; i; i = e2[i].next) {int v = e2[i].to;if(find(u) != find(v)) {vv[u].push_back(v);}}if(u == 1) return ; // printf("u : %d\n", u); // for(int i = 0; i < vv[u].size(); i++) { // printf("#%d ", vv[u][i]); // }puts("");sum = vv[u].size();ans = min(ans, sum); }int main() {int t;int m;scanf("%d",&t);for(int kase = 1; kase <= t; kase++) {scanf("%d%d",&n, &m);for(int i = 1; i <= n; i++) {vv[i].clear();}memset(num, 0, sizeof(num));memset(vis1, 0, sizeof(vis1));memset(vis2, 0, sizeof(vis2));memset(head1, 0, sizeof(head1));memset(head2, 0, sizeof(head2));tot1 = 1;tot2 = 1;int u, v;int xx;for(int i = 1; i <= n -1; i ++) {scanf("%d%d",&u, &v);add1(u, v);add1(v, u);add2(u, v);add2(v, u);}for(int i = n; i <= m; i++) {scanf("%d%d",&u, &v);add2(u, v);add2(v, u);}for(int i = 0; i <= n; i++) {fa[i] = i;}ans = 1000000000;dfs(1);printf("Case #%d: %d\n", kase, ans);}return 0; }

總結

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