hdu 2196 Computer

Computer

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31377    Accepted Submission(s): 4062

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

分析

树形dp，

dp[u][0] ： 以u为根的子树中最远的点是谁

dp[u][1] ： 以u为根的子树中次远的点是谁（那么它和dp[u][0]不是一个点）

dp[u][2] ： 通过u的父亲到达u的最远的点是谁

code

#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;

const int MAXN = 10010;
int dp[MAXN][3];

struct Edge{
    int to,nxt,w;
}e[20010];
int head[20010],tot;

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

inline void init() {
    memset(head,0,sizeof(head));
    memset(dp,0,sizeof(dp));
    tot = 0;
}

inline void add_edge(int u,int v,int w) {
    e[++tot].to = v,e[tot].w = w,e[tot].nxt = head[u],head[u] = tot; 
}

void dfs1(int u) {
    for (int i=head[u]; i; i=e[i].nxt) {
        int v = e[i].to;
        dfs1(v); // 叶 -> 根 
        int w = e[i].w;
        int tmp = dp[v][0]+w;
        if (tmp>=dp[u][0]) {
            dp[u][1] = dp[u][0];
            dp[u][0] = tmp;
        }
        else if (tmp>dp[u][1]) {
            dp[u][1] = tmp;
        }
    }
}

void dfs2(int u) {
    for (int i=head[u]; i; i=e[i].nxt) {
        int v = e[i].to,w = e[i].w;
        if (dp[u][0]==dp[v][0]+w) {
            dp[v][2] = max(dp[u][2],dp[u][1]) + w;
        }
        else {
            dp[v][2] = max(dp[u][2],dp[u][0]) + w;
        }
        dfs2(v); // 根 -> 叶 
    }
}

int main() {
    int n;
    while (scanf("%d",&n)!=EOF) {
        init();
        for (int a,b,i=2; i<=n; ++i) {
            a = read(),b = read();//表示a连向了i，权值为b。 
            add_edge(a,i,b);
        }
        dfs1(1);
        dfs2(1);
        for (int i=1; i<=n; ++i) {
            printf("%d
",max(dp[i][0],dp[i][2]));
        }
    }
    return 0;
} 

另一种做法

后来发现了一种比较简单的方法，不需要dp，

根据性质:树上任意某个节点到树上任意节点的最远距离的端点一定会是树上直径的两个端点之一

那么求出树的直径，然后在处理一下就好了，题解代码网上都有