【杭电】[2196]Computer

Computer

Time Limit: 1000/1000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)

Problem Description

A school bought the first computer some time ago(so this computer's id is 1). During the recent years the school bought N-1 new computers. Each new computer was connected to one of settled earlier. Managers of school are anxious about slow functioning of the net and want to know the maximum distance Si for which i-th computer needs to send signal (i.e. length of cable to the most distant computer). You need to provide this information. 

Hint: the example input is corresponding to this graph. And from the graph, you can see that the computer 4 is farthest one from 1, so S1 = 3. Computer 4 and 5 are the farthest ones from 2, so S2 = 2. Computer 5 is the farthest one from 3, so S3 = 3. we also get S4 = 4, S5 = 4. 

Input

Input file contains multiple test cases.In each case there is natural number N (N<=10000) in the first line, followed by (N-1) lines with descriptions of computers. i-th line contains two natural numbers - number of computer, to which i-th computer is connected and length of cable used for connection. Total length of cable does not exceed 10^9. Numbers in lines of input are separated by a space.

Output

For each case output N lines. i-th line must contain number Si for i-th computer (1<=i<=N).

 

Sample Input

5 1 1 2 1 3 1 1 1 

Sample Output

3 2 3 4 4

求树中所有点的最远距离

先找任意点的最远点u，在找u的最远点v，在这个过程中记录u到每个点的距离dis[0][i]，再以v为起点寻找最远点，记录v到每个点的距离dis[1][i]，则各个点的最远距离为max(dis[0][i],dis[1][i])

#include<stdio.h>
#include<string.h>
int n,m;
int t,max;
int dis[2][10200];
int head[10200];
int headcnt;
struct List {
	int u,v,w;
	int next;
} edge[20200];
int f(int a,int b) {
	return a>b?a:b;
}
void dfs(int u,int v,int w,int flag) {
	if(max<w)
		max=w,t=u;
	dis[flag][u]=w;
	for(int i=head[u]; i!=-1; i=edge[i].next) {
		if(edge[i].v!=v) {
			dfs(edge[i].v,u,w+edge[i].w,flag);
		}
	}
}
void add(int u,int v,int w) {
	edge[headcnt].u=u;
	edge[headcnt].v=v;
	edge[headcnt].w=w;
	edge[headcnt].next=head[u];
	head[u]=headcnt++;
}
int main() {
	while(scanf("%d",&n)!=EOF) {
		headcnt=0;
		memset(head,-1,sizeof(head));
		for(int i=2; i<=n; i++) {
			int v,w;
			scanf("%d %d",&v,&w);
			add(i,v,w);
			add(v,i,w);
		}
		max=0;
		dfs(1,0,0,0);
		max=0;
		dfs(t,0,0,0);
		dfs(t,0,0,1);
		for(int i=1; i<=n; i++) {
			printf("%d
",f(dis[0][i],dis[1][i]));
		}
	}
	return 0;
}

题目地址:【杭电】[2196]Computer