hdu 2196 Computer(樹形DP)

標籤：des   style   blog   http   java   color   os   io   

Computer Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3075    Accepted Submission(s): 1561

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

51 12 13 11 1

Sample Output

32344

::所有電腦串連成樹形結構，要求求出每個結點到離該結點最遠距離。

如果以每個結點為根結點，進行一次dfs，那就很容易求出那個最遠距離。但是如果只是單純的暴力，時間是不夠的，

不過我們可以用DP的思想，避免計算重複的子問題，利用這個剪枝，效率就相當高。

我的實現方法：以每個結點為根結點root，進行一次dfs，用dp儲存的是經過某條邊能到達的最遠距離。

邊的儲存是雙向的，給每條邊一個序號。//下面dp[x], x就是邊的序號

以為例，首先選取任意一結點這裡選1，為根結點，進行dfs，然後我們就可以求出經過邊2邊1..邊4….所能到達的最遠距離dp[1], d[2]……dp[4]..

然後以2為根結點dfs, 因為經過邊4所能到達最遠距離dp[4]以知道。。dp[7], dp[2]也知道，那離結點2最遠結點的距離等於max(dp[4], dp[7], dp[2]+dis[3]),//dis[3]表示第三條邊的長度

view code#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 10010;int n, dp[N<<1], pre[N];struct edge{    int u, v, w, p;    edge() {}    edge(int u, int v, int w, int p):u(u), v(v), w(w), p(p) {}}e[N<<1];int ecnt = 0;int dfs(int u, int p){    int ans = 0;    for(int i=pre[u]; ~i; i=e[i].p)    {        int v = e[i].v;        if(v==p) continue;        if(!dp[i]) dp[i] = dfs(v, u)+e[i].w;        ans = max(ans , dp[i]);    }    return ans;}int main(){//    freopen("in", "r", stdin);    while(scanf("%d", &n)>0)    {        memset(dp, 0, sizeof(dp));        memset(pre, -1, sizeof(pre));        int v, w;        ecnt = 0;        for(int i=2; i<=n; i++)        {            scanf("%d%d", &v, &w);            e[ecnt] = edge(i, v, w, pre[i]);            pre[i] = ecnt++;            e[ecnt] = edge(v, i, w, pre[v]);            pre[v] = ecnt++;        }        for(int i=1; i<=n; i++)            printf("%d\n", dfs(i,-1));    }    return 0;}