HDOJ 2196 Computer

经典的树形DP。。。

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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.

 

 5
1 1
2 1
3 1
1 1 

 3
2
3
4
4 

Source

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

using namespace std;

const int maxn=22000;
const int INF=0x3f3f3f3f;

int Adj[maxn],Size;

struct E
{
	int to,next,len;
}Edge[maxn];

void init()
{
	Size=0;
	memset(Adj,-1,sizeof(Adj));
}

void Add_Edge(int u,int v,int w)
{
	Edge[Size].to=v;
	Edge[Size].len=w;
	Edge[Size].next=Adj[u];
	Adj[u]=Size++;
}

int n,dp[maxn][3];

void dfs1(int f,int u)
{
    int biggest=0,bigger=0;
    for(int i=Adj[u];~i;i=Edge[i].next)
    {
        int v=Edge[i].to;
        int w=Edge[i].len;
        if(v==f) continue;
        dfs1(u,v);
        int temp=dp[v][0]+w;
        if(temp>=biggest)
        {
            bigger=biggest;
            biggest=temp;
        }
        else if(temp>bigger)
        {
            bigger=temp;
        }
    }
    dp[u][0]=biggest;
    dp[u][1]=bigger;
}

void dfs2(int f,int u)
{
    for(int i=Adj[u];~i;i=Edge[i].next)
    {
        int v=Edge[i].to;
        int w=Edge[i].len;
        if(v==f) continue;
        dp[v][2]=max(dp[u][2],(dp[v][0]+w==dp[u][0])?dp[u][1]:dp[u][0]) + w ;
        dfs2(u,v);
    }
}

int main()
{
while(scanf("%d",&n)!=EOF)
{
	init();
	for(int i=2;i<=n;i++)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		Add_Edge(i,a,b);
		Add_Edge(a,i,b);
	}
	memset(dp,0,sizeof(dp));
    dfs1(-1,1);
    dfs2(-1,1);
	for(int i=1;i<=n;i++)
    {
        printf("%d\n",max(dp[i][0],dp[i][2]));
    }
}
	return 0;
}

HDOJ 2196 Computer

原文：http://blog.csdn.net/u012797220/article/details/18940215