POJ3107 Godfather (树形DP)

题意:求树的重心

题解:先跑一遍dfs 预处理出这种遍历方式每个节点的儿子(含自己)的数

　　 再跑一遍 每个点的值就是他所有儿子中取一个最大值 再和它父亲这个方向比较一下

　　 又被卡常了 vector一直tle 需要用前向星....

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <vector>
#include <string.h>
using namespace std;

int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

struct node
{
    int no, to, nex;
}E[100005];

int n, rt;
int du[50005];
int head[50005];
int son[50005];
int dp[50005];

int dfs1(int x, int fa)
{
    son[x] = 1;
    int c = head[x];
    for(int i = c; i; i = E[i].nex)
    {
        int c = E[i].to;
        if(c == fa) continue;

        son[x] += dfs1(c, x);
    }
    return son[x];
}

void dfs2(int x, int fa)
{
    dp[x] = n - son[x];
    int c = head[x];
    for(int i = c; i; i = E[i].nex)
    {
        int c = E[i].to;
        if(c == fa) continue;

        dp[x] = max(dp[x], son[c]);
        dfs2(c, x);
    }
}

int main()
{
    while(~scanf("%d", &n))
    {
        for(int i = 1; i <= n; i++) du[i] = son[i] = dp[i] = head[i] = 0;

        int cn = 0;
        for(int i = 1; i < n; i++)
        {
            int u, v;
            u = read(); v = read();
            E[++cn].no = u;
            E[cn].to = v;
            E[cn].nex = head[u];
            head[u] = cn;

            E[++cn].no = v;
            E[cn].to = u;
            E[cn].nex = head[v];
            head[v] = cn;
            du[u]++;
            du[v]++;
        }


        for(int i = 1; i <= n; i++)
            if(du[i] == 1)
            {
                rt = i;
                break;
            }

        int o = dfs1(rt, -1);
        dfs2(rt, -1);

        int ans = n + 5;
        int cnt = 0;
        for(int i = 1; i <= n; i++) ans = min(ans, dp[i]);
        for(int i = 1; i <= n; i++)
            if(dp[i] == ans) cnt++;

        for(int i = 1; i <= n; i++)
        {
            if(dp[i] == ans)
            {
                if(cnt == 1)
                {
                    printf("%d
", i);
                    break;
                }
                else printf("%d ", i);
                cnt--;
            }
        }
    }
    return 0;
}