无向图缩点poj3177 Redundant Paths

Posted 2020-10-27 kang000

http://poj.org/problem?id=3177

题意：有n个牧场，互相连通，现在要求使得任意两个牧场之间至少有两条道路可走，求至少需要修多少条新的路

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求出双连通分量中，度为1的个数ans，答案即为（ans+1）/2
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 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <algorithm>
 5 #include <stack>
 6 using namespace std;
 7 const int N = 1e5 + 50;
 8 int vis[N], low[N], dfn[N], cnt[N], belong[N];
 9 bool mp[5050][5050];
10 int E, deep, n, m, cnt2;
11 stack<int>stk;
12 void dfs(int u, int fa)
13 {
14     vis[u] = 1;  //在栈中
15     stk.push(u);
16     low[u] = dfn[u] = ++deep;
17     for (int i = 1; i <= n; i++)
18     {
19         if (mp[u][i] && i != fa)
20         {
21             if (!vis[i]){
22                 dfs(i, u);
23                 low[u] = min(low[u], low[i]);
24             }
25             else
26                 low[u] = min(low[u], dfn[i]);
27         }
28     }
29     if (low[u] == dfn[u] && vis[u]){
30         cnt2++;  //双连通分量个数
31         while (stk.top() != u){
32             vis[stk.top()] = 0;
33             belong[stk.top()] = cnt2;
34             stk.pop();
35         }
36         vis[u] = 0;
37         belong[u] = cnt2;
38         stk.pop();
39     }
40 }
41 int main(){
42     while (scanf("%d%d", &n, &m) != EOF)
43     {
44         E = deep = 0;
45         memset(low, 0, sizeof(low));
46         memset(dfn, 0, sizeof(dfn));
47         memset(vis, 0, sizeof(vis));
48         memset(cnt, 0, sizeof(cnt));
49         memset(belong, 0, sizeof(belong));
50         while (stk.size())stk.pop();
51         cnt2 = 0;
52         for (int i = 1; i <= m; i++)
53         {
54             int u, v;
55             scanf("%d%d", &u, &v);
56             mp[u][v] = mp[v][u] = 1;
57         }
58         dfs(1, 1);
59         for (int i = 1; i <= n; i++)
60             for (int j = 1; j <= n; j++)
61                 if (mp[i][j] && belong[i] != belong[j]) //遍历所有边，度++
62                     cnt[belong[i]]++, cnt[belong[j]]++, mp[i][j] = mp[j][i] = 0;
63         int ans = 0;
64         for (int i = 1; i <= cnt2; i++)
65             if (cnt[i] == 1) ans++;
66         cout << (ans + 1) / 2 << endl;
67     }
68 }