Computer

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.

5
1 1
2 1
3 1
1 1

3
2
3
4
4

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<vector>
using namespace std;

typedef long long ll;
#define M 10005

vector<int>  adj[M],wi[M];
ll f[M][2];
int vis[M];

ll dfs1(int u){
    vis[u]=1;
    f[u][0]=0;

    for(int i=0;i<adj[u].size();i++){
        int v=adj[u][i];
        int w=wi[u][i];
        if(vis[v])   continue;

        f[u][0]=max(f[u][0],dfs1(v)+w);
    }
    return f[u][0];
}

void dfs2(int u){
    vis[u]=1;
    int m1=0,m2=0,v1,v2;

    for(int i=0;i<adj[u].size();i++){
        int v=adj[u][i];
        int w=wi[u][i];
        if(vis[v])  continue;

        int tmp=f[v][0]+w;
        if(tmp>m1){
            m2=m1,v2=v1;
            m1=tmp,v1=v;
        }
        else  if(m1==tmp || tmp>m2){
            m2=tmp,v2=v;
        }
    }

    if(u!=1){
        int tmp=f[u][1];
        int v=-1;
        if(tmp>m1){
            m2=m1,v2=v1;
            m1=tmp,v1=v;
        }
        else  if(m1==tmp || tmp>m2){
            m2=tmp,v2=v;
        }
    }

    for(int i=0;i<adj[u].size();i++){
        int v=adj[u][i];
        int w=wi[u][i];
        if(vis[v])  continue;

        if(v==v1)  f[v][1]=m2+w;
        else       f[v][1]=m1+w;
        dfs2(v);
    }
}

int main(){
    int n;
    while(cin>>n && n){
        for(int i=0;i<=n;i++)
            adj[i].clear(),wi[i].clear();
        for(int u=2;u<=n;u++){
            int v,w;
            cin>>v>>w;
            adj[u].push_back(v);
            wi[u].push_back(w);
            adj[v].push_back(u);
            wi[v].push_back(w);
        }

        memset(f,0,sizeof f);
        memset(vis,0,sizeof vis);
        dfs1(1);
        memset(vis,0,sizeof vis);
        dfs2(1);

        for(int j=1;j<=n;j++)
            cout<<max(f[j][0],f[j][1])<<endl;
    }
}

hdu 2196 Computer DP

原文：http://blog.csdn.net/hyccfy/article/details/42914533