POJ 1655 Balancing Act (dfs)

题意：给定一棵树，问拿掉那个点，能使余下的各个子树结点个数上限最小。

看了半天，只知道是深搜，可是想了很久也没想出实现方法，各种菜~~~~看了大牛的实现，才恍然大悟........dfs的灵活之处自己需要时间去掌握啊.

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <memory.h>
#include <cmath>
#include <set>
#include <queue>
#include <vector>
using namespace std;

const int BORDER = (1<<20)-1;
#define MAXN 20020
#define INF 0x7ffffff
#define CLR(x,y) memset(x,y,sizeof(x))
#define ADD(x) ((++x)&BORDER)
#define IN(x) scanf("%d",&x)
#define OUT(x) printf("%d\n",x)
#define MIN(m,v) (m)<(v)?(m):(v)
#define MAX(m,v) (m)>(v)?(m):(v)
#define ABS(x) (x>0?x:-x)

struct EDGE{
    int v,next;
}edge[MAXN*2];
int n,m,cnt,mmin_index,mmin_cnt,index;
int net[MAXN],visit[MAXN];
void add_edge(const int& u,const int& v)
{
    edge[index].v = v;
    edge[index].next = net[u];
    net[u] = index;
    ++index;
    edge[index].v = u;
    edge[index].next = net[v];
    net[v] = index;
    ++index;
}
int init()
{
    index = 0;
    mmin_index = INF;
    mmin_cnt = INF;
    CLR(net,-1);
    CLR(visit,0);
    return 0;
}
int dfs(const int& u)
{
    int i,v;
    int mmax = 0;
    int sum = 0;
    visit[u] = 1;
    for(i = net[u]; i != -1; i = edge[i].next)
    {
        v = edge[i].v;
        if(!visit[v])
        {
            int tmp = dfs(v);
            mmax = MAX(mmax,tmp);
            sum += tmp;
        }
    }
    mmax = MAX(mmax,n-sum-1);
    if(mmin_cnt >= mmax)
    {
        if(mmin_cnt == mmax && mmin_index < u)
            ;
        else
        {
            mmin_cnt = mmax;
            mmin_index = u;
        }
    }
    return sum+1;
}
int input()
{
    int i,j,tmp,a,b;
    IN(n);

    tmp = n - 1;
    for(i = 0; i < tmp; ++i)
    {
        scanf("%d %d",&a,&b);
        add_edge(a,b);
    }
    return 0;
}
int work()
{
    dfs(1);
    printf("%d %d\n",mmin_index,mmin_cnt);
    return 0;
}

int main()
{
    int i,j,tt,tmp;
    IN(tt);
    while(tt--)
    {
        init();
　　　　 input();

        work();
    }
    return 0;
}