Computer

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7329 Accepted Submission(s): 3630

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

Sample Output

题意是求树中每个点到所有叶子节点的距离的最大值是多少。

//经典树形DP
//进行第一次dfs,先以1为根，求出每个节点以该节点为根的子树的最大值和第二大的值
//进行第二次dfs, 对于每个节点v的最大值与第二大值，可能是来自自己的子树或者父节点来的
//判断它来自哪，这取决于该节点v父亲fa[v]取的最大值是来自哪个节点book[v]
//如果book[v]==v，说明fa[v]取的最大值来源于节点v,所以v节点的最大值来源不能是fa[v]的最大值,只能来源于fa[v]第二大或者它的子节点
//如果book[v]!=v, 说明fa[v]取的最大值不是来源于节点v,所以v节点的最大值来源可以是fa[v]的最大值也可以是它的子节点，找最大的
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#define maxn 10005
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;

struct node
{
	int v, w;
};

vector<node>G[maxn];
ll dp[maxn][2];                     //0是最大，1是第二大
int fa[maxn];                       //父亲
int book[maxn];                     //book[i]表示节点i的最大距离的路径经过的第一个点
//ll num[maxn];//存最终结果
//int vis[maxn];

void dfs1(int root)
{
	int l = G[root].size();
	for(int i=0; i<l; i++)
	{
		int to = G[root][i].v;
		dfs1(to);
	}

	//找第一大
	int num1 = 0;
	for(int i=0; i<l; i++)
	{
		node to = G[root][i];
		if(dp[root][0] < dp[to.v][0]+to.w)
		{
			num1 = i;
			dp[root][0] = dp[to.v][0]+to.w;
			book[root] = to.v;
		}
	}
	//找第二大
	for(int i=0; i<l; i++)
	{
		node to = G[root][i];
		if(dp[root][1] < dp[to.v][0]+to.w && num1!=i)
		{
			dp[root][1] = dp[to.v][0]+to.w;
		}
	}

}

void dfs2(int root, int cnt)            //cnt表示root是它父节点的第cnt个孩子
{
    if(root!=1)
    {
    if(book[fa[root]]!=root)
    {
        if(dp[fa[root]][0]+G[fa[root]][cnt].w > dp[root][0])
        {
            dp[root][1] = dp[root][0];
            dp[root][0] = dp[fa[root]][0]+G[fa[root]][cnt].w;
            //printf("fsafa== %d\n", dp[root][0]);
            book[root] = fa[root];
        }
        else
        {
            dp[root][1] = max(dp[root][1],dp[fa[root]][0]+G[fa[root]][cnt].w);
        }
    }
    else
    {
        if(dp[fa[root]][1]+G[fa[root]][cnt].w > dp[root][0])
        {
            dp[root][1] = dp[root][0];
            dp[root][0] = dp[fa[root]][1]+G[fa[root]][cnt].w;
            book[root] = fa[root];
        }
        else
        {
            dp[root][1] = max(dp[root][1],dp[fa[root]][1]+G[fa[root]][cnt].w);
        }
    }
    }
    int l = G[root].size();
    for(int i=0; i<l; i++)
        dfs2(G[root][i].v, i);
}
/*
void dfs3(int root, int cnt)          //cnt表示root是它父节点的第cnt个孩子
{
	int l = G[root].size();
	for(int i=0; i<l; i++)
	{
		int to = G[root][i].v;
		dfs3(to, i);
	}
    if(root==1)
    {
        num[root] = dp[root][0];
        return;
    }
    if(book[fa[root]]!=root)
    {
        num[root] = max(dp[root][0], dp[fa[root]][0]+G[fa[root]][cnt].w);
    }
    else
    {
        num[root] = max(dp[root][0], dp[fa[root]][1]+G[fa[root]][cnt].w);
    }
}
*/
int main()
{
	int n;
	while(~scanf("%d", &n))
	{
		for(int i=0; i<maxn; i++)
        {
            fa[i] = 0;
            book[i] = 0;
            num[i] = 0;
            dp[i][0] = dp[i][1] = 0;
            G[i].clear();
        }
		for(int i=0; i<n-1; i++)
		{
			int u, w;
			scanf("%d%d", &u, &w);
			G[u].push_back((node){i+2, w});
			//G[i+2].push_back((node){u, w});
			fa[i+2] = u;
		}
//		printf("yes\n");
		dfs1(1);
//		printf("yes1\n");
		dfs2(1,0);
//		printf("yes2\n");
//		dfs3(1,0);
//		printf("yes3\n");
		for(int i=1; i<=n; i++)
		{
		   // printf("dp[%d][0] = %lld\n", i, dp[i][0]);
		    //printf("dp[%d][1] = %lld\n", i, dp[i][1]);
		    //printf("book[%d] = %d\n", i, book[i]);
			printf("%lld\n", dp[i][0]);
		}
	}
	return 0;
}

HDU 2196-Computer(经典树形DP)

Computer

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