Hdu 5452 Minimum Cut (2015 ACM/ICPC Asia Regional Shenyang Online) dfs + LCA

题目链接：

　　Hdu 5452 Minimum Cut

题目描述：

　　有一棵生成树，有n个点，给出m-n+1条边，截断一条生成树上的边后，再截断至少多少条边才能使图不连通， 问截断总边数？

解题思路：

　　因为只能在生成树上截断一条边(u, v)，所以只需要统计以v为根节点的子生成树里的节点与子生成树外的节点的边数就可以了。对于新加入的边(u‘, v‘)来说，只影响以LCA(u, v)为根节点的子树里面的节点。统计所有答案，扫一遍输出最小即可。(比赛的时候只统计叶子节点，给水过去了........23333333)

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 using namespace std;
 6
 7 const int INF = 0x3f3f3f3f;
 8 const int maxn = 20010;
 9 struct node
10 {
11     int to, next;
12 }edge[maxn*2];
13 int tot, head[maxn], deep[maxn], fa[maxn], ans[maxn];
14
15 void init ()
16 {
17     tot = 0;
18     memset (head, -1, sizeof(head));
19 }
20
21 void Add (int from, int to)
22 {
23     edge[tot].to = to;
24     edge[tot].next = head[from];
25     head[from] = tot ++;
26 }
27
28 void dfs (int u, int father, int dep)
29 {
30     deep[u] = dep;
31     ans[u] = 1;
32     fa[u] = father;
33
34     for (int i=head[u]; i!=-1; i=edge[i].next)
35     {
36         int v = edge[i].to;
37         if (v != father)
38             dfs (v, u, dep+1);
39     }
40 }
41
42 void LCA (int x, int y)
43 {
44     while (x != y)
45     {
46        if (deep[x] >= deep[y])
47        {
48            ans[x] ++;
49            x = fa[x];
50        }
51        else
52        {
53            ans[y] ++;
54            y = fa[y];
55        }
56     }
57 }
58
59 int main ()
60 {
61     int t, n, m, i;
62     scanf ("%d", &t);
63
64     for (int j=1; j<=t; j++)
65     {
66         init ();
67         int u, v;
68         scanf ("%d %d", &n, &m);
69
70         for (i=1; i<n; i++)
71         {
72             scanf ("%d %d", &u, &v);
73             Add (u, v);
74             Add (v, u);
75         }
76         dfs (1, 0, 1);
77
78         for ( ;i<=m; i++)
79         {
80             scanf ("%d %d", &u, &v);
81             LCA (u, v);
82         }
83
84         int res = INF;
85         for (i=2; i<=n; i++)
86             res = min (res, ans[i]);
87         printf ("Case #%d: %d\n", j, res);
88     }
89     return 0;
90 }