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)

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int INF = 0x3f3f3f3f;
 const int maxn = ;
 struct node
 {
     int to, next;
 }edge[maxn*];
 int tot, head[maxn], deep[maxn], fa[maxn], ans[maxn];

 void init ()
 {
     tot = ;
     memset (head, -, sizeof(head));
 }

 void Add (int from, int to)
 {
     edge[tot].to = to;
     edge[tot].next = head[from];
     head[from] = tot ++;
 }

 void dfs (int u, int father, int dep)
 {
     deep[u] = dep;
     ans[u] = ;
     fa[u] = father;

     for (int i=head[u]; i!=-; i=edge[i].next)
     {
         int v = edge[i].to;
         if (v != father)
             dfs (v, u, dep+);
     }
 }

 void LCA (int x, int y)
 {
     while (x != y)
     {
        if (deep[x] >= deep[y])
        {
            ans[x] ++;
            x = fa[x];
        }
        else
        {
            ans[y] ++;
            y = fa[y];
        }
     }
 }

 int main ()
 {
     int t, n, m, i;
     scanf ("%d", &t);

     for (int j=; j<=t; j++)
     {
         init ();
         int u, v;
         scanf ("%d %d", &n, &m);

         for (i=; i<n; i++)
         {
             scanf ("%d %d", &u, &v);
             Add (u, v);
             Add (v, u);
         }
         dfs (, , );

         for ( ;i<=m; i++)
         {
             scanf ("%d %d", &u, &v);
             LCA (u, v);
         }

         int res = INF;
         for (i=; i<=n; i++)
             res = min (res, ans[i]);
         printf ("Case #%d: %d\n", j, res);
     }
     return ;
 }