hdu 2196 Computer（DP-树形DP）

Computer

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3152    Accepted Submission(s): 1598

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

 

每个节点记录，当前节点到它子树的最远距离和次远距离，以及这两个距离分别来自哪个儿子，这是第一遍DP做的。

第二遍DP求出每个点到所有点的最远距离。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define ll long longconst int maxn = 10010;struct Node{    ll Maxlen[2] , son[2];}dp[maxn];struct edge{    int u , v , cost , next;    edge(int a =0 , int b = 0 , int c = 0 , int d = 0){        u = a , v = b , cost = c , next = d;    }}e[4*maxn];int N , cnt , head[maxn] , vis[maxn];void add(int u , int v , int c){    e[cnt] = edge(u , v , c , head[u]);    head[u] = cnt++;    e[cnt] = edge(v , u , c , head[v]);    head[v] = cnt++;}void initial(){    for(int i = 0; i < maxn; i++){        head[i] = -1;        dp[i].Maxlen[0] = 0;        dp[i].Maxlen[1] = 0;        dp[i].son[0] = -1;        dp[i].son[1] = -1;        vis[i] = 0;    }    cnt = 0;}void readcase(){    for(int u = 2 ; u <= N ; u++){        int v , c;        scanf("%d%d" , &v , &c);        add(u , v , c);    }}void DP1(int u){    vis[u] = 1;    for(int i = head[u]; i != -1; i = e[i].next){        int v = e[i].v;        int c = e[i].cost;        if(!vis[v]){            DP1(v);            if(dp[v].Maxlen[0]+c>dp[u].Maxlen[0]){                dp[u].Maxlen[1] = dp[u].Maxlen[0];                dp[u].son[1] = dp[u].son[0];                dp[u].Maxlen[0] = dp[v].Maxlen[0]+c;                dp[u].son[0] = v;            }else if(dp[v].Maxlen[0]+c>dp[u].Maxlen[1]){                dp[u].Maxlen[1] = dp[v].Maxlen[0]+c;                dp[u].son[1] = v;            }        }    }}void DP2(int u , int f , int cost){    vis[u] = 1;    if(dp[f].son[0] != u && dp[f].Maxlen[0]+cost >= dp[u].Maxlen[0]){        dp[u].Maxlen[0] = dp[f].Maxlen[0]+cost;        dp[u].son[0] = f;    }else if(dp[f].son[1] != u && dp[f].Maxlen[1]+cost >= dp[u].Maxlen[0]){        dp[u].Maxlen[0] = dp[f].Maxlen[1]+cost;        dp[u].son[0] = f;    }else if(dp[f].son[0] != u && dp[f].Maxlen[0]+cost >= dp[u].Maxlen[1]){        dp[u].Maxlen[1] = dp[f].Maxlen[0]+cost;        dp[u].son[1] = f;    }else if(dp[f].son[1] != u && dp[f].Maxlen[1]+cost >= dp[u].Maxlen[1]){        dp[u].Maxlen[1] = dp[f].Maxlen[1]+cost;        dp[u].son[1] = f;    }    //cout << "u=" << u << " f=" << f << " cost=" << cost << " Max=" << dp[u].Maxlen[1] << " son=" << dp[u].son[1] << endl;    for(int i = head[u]; i != -1; i = e[i].next){        int v = e[i].v;        int c = e[i].cost;        if(!vis[v]) DP2(v , u , c);    }}void computing(){    DP1(1);    memset(vis , 0 , sizeof vis);    vis[0] = 1;    DP2(1 , 0 , 0);    for(int i = 1; i <= N; i++){        cout << dp[i].Maxlen[0] << endl;    }}int main(){    while(~scanf("%d" , &N)){        initial();        readcase();        computing();    }    return 0;}