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其实很简单，先计算子树的max，再计算父树的max就OK了；

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#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
#define rep(i,n) for(int i=1;i<=n;i++)
#define clr(x,c) memset(x,c,sizeof(x))

int read(){
 int x=0； int f=1;
 char c=getchar();
 while(!isdigit(c)){
  if(c==‘-‘) f=-1;
  c=getchar();
 }
 while(isdigit(c)){
  x=x*10+c-‘0‘;
  c=getchar();
 }
 return x*f;
}

struct edge{
 int to,w;
 edge*next;
};
edge e[10005],*pt,*head[10005];int f[10005][3];

void add(int s,int t,int w){
 pt->to=t; pt->w=w;
 pt->next=head[s];
 head[s]=pt++;
}
void dp1(int x){
 for(edge*ee=head[x];ee;ee=ee->next){
  int to=ee->to; int w=ee->w;
  dp1(to);
  if(f[to][1]+w>f[x][1]){
   f[x][0]=f[x][1];f[x][1]=f[to][1]+w;
  }
  else if(f[to][1]+w>f[x][0]){
   f[x][0]=f[to][1]+w;
  }
 }
 return ;
}
void dp2(int x,int fa,int len){
 if(f[x][1]+len==f[fa][1]){
  f[x][2]=max(f[fa][2],f[fa][0]);
 }
 else{
  f[x][2]=max(f[fa][2],f[fa][1]);
 }
 f[x][2]+=len;
 for(edge*ee=head[x];ee;ee=ee->next){
  int to=ee->to;int w=ee->w;
  dp2(to,x,w);
 }
 return ;
}
int main(){
 int n;
 while(scanf("%d",&n)==1){
  clr(f,0);clr(head,0);
  pt=e;
  for(int i=2;i<=n;i++){
   int u=read(),v=read();
   add(u,i,v);
  }
  /*rep(i,n){
      cout<<i<<endl;
      for(edge*ee=head[i];ee;ee=ee->next)
       printf("%d %d\n",ee->to,ee->w); 

 }*/
  dp1(1);
  dp2(1,0,0);
  rep(i,n)   printf("%d\n",max(f[i][1],f[i][2]));
 }
 return 0;
}

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Computer

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4926    Accepted Submission(s): 2475

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.

hdu2196：树形dp

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原文地址：http://www.cnblogs.com/fighting-to-the-end/p/5049554.html