Minimum Cut
Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 769 Accepted Submission(s): 340
Problem Description
Given a simple unweighted graph G (an undirected graph containing no loops nor multiple edges) with n nodes and m edges. Let T be a spanning tree of G.
We say that a cut in G respects T if it cuts just one edges of T.
Since love needs good faith and hypocrisy return for only grief, you should find the minimum cut of graph G respecting the given spanning tree T.
Input
The input contains several test cases.
The first line of the input is a single integer t (1≤t≤5) which is the number of test cases.
Then t test cases follow.
Each test case contains several lines.
The first line contains two integers n (2≤n≤20000) and m (n−1≤m≤200000).
The following n−1 lines describe the spanning tree T and each of them contains two integers u and v corresponding to an edge.
Next m−n+1 lines describe the undirected graph G and each of them contains two integers u and v corresponding to an edge which is not in the spanning tree T.
Output
For each test case, you should output the minimum cut of graph G respecting the given spanning tree T.
Sample Input
1
4 5
1 2
2 3
3 4
1 3
1 4
Sample Output
Case #1: 2
Source
2015 ACM/ICPC Asia Regional Shenyang Online
题意：一无权无向无环无多余边图G，由n个结点，m条边构成。T是其生成树。如果T除去一条边，G有一个割可以代表T，我们这么说。找最小割
输入n， m；前 n -1行代表T，之后m-n+1行是G中T没有的边。求最小割。
割就是，在T中删除一条边，在G中需要删除几条边可以变成T的图的连接关系。在m-n-1行中都是T中没有的边，添加了（u， v）之后，v中的子节点，叶子节点都可以和u联系，那么在求最小割的时候就要删除这条边，使得v中原本不能与u联系的还是不能联系变成 T 图的关系。
用cnt存该点由G变成T（在T中删除一条边）变 了 n-1次（一次删除一条边）之后，需要被删除的次数。遍历求最小值即最小割
1 #include <iostream>
2 #include <cstdlib>
3 #include <cstdio>
4 #include <algorithm>
5 #include <vector>
6 #include <queue>
7 #include <cmath>
8 #include <stack>
9 #include <cstring>
10
11 using namespace std;
12
13 #define INF 0x3f3f3f3f
14 #define min(a,b) (a<b?a:b)
15 #define N 100005
16
17 vector< vector<int> > G;
18
19 int n, m, cnt[N], dfn[N], f[N];
20
21 void init()
22 {
23 G.resize(N);
24 G.clear();
25 memset(dfn, 0, sizeof(dfn));
26 }
27
28 void Tarjan(int u, int fa, int step)
29 {
30 dfn[u] = step;
31 cnt[u] = 1;
32 f[u] = fa;
33 int len = G[u].size();
34 for(int i = 0; i < len; i++)
35 {
36 int v = G[u][i];
37 if(v != fa)
38 Tarjan(v, u, step+1);
39 }
40 }
41
42 void Lca(int x, int y)
43 {
44 while(x != y)
45 {
46 if(dfn[x] >= dfn[y]) //dfn表示树的深度，找深的父节点
47 {
48 cnt[x]++;
49 x = f[x];
50 }
51 else
52 {
53 cnt[y]++;
54 y = f[y];
55 }
56 }
57 }
58
59 int main()
60 {
61 int t, i, a, b, l = 1;
62 scanf("%d", &t);
63 while(t--)
64 {
65 init();
66 scanf("%d%d", &n, &m);
67 for(i = 1; i < n; i++)
68 {
69 scanf("%d%d", &a, &b);
70 G[a].push_back(b);
71 G[b].push_back(a);
72 }
73 Tarjan(1, 0, 1);
74 for(; i <= m; i++)
75 {
76 scanf("%d%d", &a, &b);
77 Lca(a, b);
78 }
79 int ans = INF;
80 for(i = 2; i <= n; i++)
81 ans = min(ans, cnt[i]);
82 printf("Case #%d: %d\n", l++ ,ans);
83 }
84 return 0;
85 }