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 = 20010;
struct node
{
    int to, next;
}edge[maxn*2];
int tot, head[maxn], deep[maxn], fa[maxn], ans[maxn];

void init ()
{
    tot = 0;
    memset (head, -1, 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] = 1;
    fa[u] = father;
    
    for (int i=head[u]; i!=-1; i=edge[i].next)
    {
        int v = edge[i].to;
        if (v != father)
            dfs (v, u, dep+1);
    }
}

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=1; j<=t; j++)
    {
        init ();
        int u, v;
        scanf ("%d %d", &n, &m);

        for (i=1; i<n; i++)
        {
            scanf ("%d %d", &u, &v);
            Add (u, v);
            Add (v, u);
        }
        dfs (1, 0, 1);
        
        for ( ;i<=m; i++)
        {
            scanf ("%d %d", &u, &v);
            LCA (u, v);
        }
        
        int res = INF;
        for (i=2; i<=n; i++)
            res = min (res, ans[i]);
        printf ("Case #%d: %d
", j, res);
    }
    return 0;
}