hdu 2196 Computer 树形dp 树中点最大距离

//	hdu 2196 Computer 树形dp
//
//	题目大意:
//
//		求树上每个点到其他点的最大距离
//
//	解题思路:
//
//		保存每个点到其子树的最大距离d[u][0]和次大距离d[u][1].对与
//	最大距离d[v][2]的求法和它的父亲u有关.如果v是在u的最大距离的子
//	树中,那么答案就是d[u][1] + dist[u,v]否则就是d[u][0] + dist[u,v]
//	哎,自己的功力还是不够,这么经典的都是最近才知道,继续加油吧!!!

#include <cstring>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <map>
#define For(x,a,b,c) for (int x = a; x <= b; x += c)
#define Ffor(x,a,b,c) for (int x = a; x >= b; x -= c)
#define cls(x,a) memset(x,a,sizeof(x))
using namespace std;
typedef long long ll;

const int MAX_N = 10008;

const int INF = 0x3f3f3f3f;
const ll MOD = 1e6;
int N,M;

struct node{
int v;
int w;

node(){

}
node(int v,int w):v(v),w(w){

}

};

vector<node> g[MAX_N];

int d[MAX_N][3];

void dfs1(int u,int fa){
int first = 0,second = 0;

for (int i = 0;i < g[u].size();i ++){
int v = g[u][i].v;

if (v == fa)
continue;

dfs1(v,u);

int tmp = g[u][i].w + d[v][0];

if (first <= tmp){
second = first;
first = tmp;
}
else if (second < tmp){
second = tmp;
}
}

d[u][0] = first;
d[u][1] = second;
}

void dfs2(int u,int fa){
for (int i = 0 ;i < g[u].size();i ++){
int v = g[u][i].v;
if (v == fa)
continue;

int w = g[u][i].w;

d[v][2] = max(d[v][2],max(d[u][2],d[v][0] + w == d[u][0] ? d[u][1] : d[u][0]) + w);
dfs2(v,u);

}
}

void input(){

for (int i = 1;i <= N ; i ++)
g[i].clear();

for (int i = 2;i <= N;i ++){
int u,v;
scanf("%d%d",&u,&v);
g[i].push_back(node(u,v));
g[u].push_back(node(i,v));
}

dfs1(1,-1);

for (int i = 1;i <= N;i ++)
d[i][2] = 0;

dfs2(1,-1);

for (int i = 1;i <= N;i ++){
printf("%d\n",max(d[i][0],d[i][2]));
}

}

int main(){
//freopen("1.in","r",stdin);
while(scanf("%d",&N)!=EOF){
input();
}
return 0;
}