HDU 4276 The Ghost Blows Light 第37届ACM/ICPC长春赛区1010题 （树形DP）

The Ghost Blows Light

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 590    Accepted Submission(s): 194

Problem Description

My name is Hu Bayi, robing an ancient tomb in Tibet. The tomb consists of N rooms (numbered from 1 to N) which are connected by some roads (pass each road should cost some time). There is exactly one route between any two rooms, and each room contains some treasures. Now I am located at the 1st room and the exit is located at the Nth room.
Suddenly, alert occurred! The tomb will topple down in T minutes, and I should reach exit room in T minutes. Human beings die in pursuit of wealth, and birds die in pursuit of food! Although it is life-threatening time, I also want to get treasure out as much as possible. Now I wonder the maximum number of treasures I can take out in T minutes.

 

 

Input

There are multiple test cases.
The first line contains two integer N and T. (1 <= n <= 100, 0 <= T <= 500)
Each of the next N - 1 lines contains three integers a, b, and t indicating there is a road between a and b which costs t minutes. (1<=a<=n, 1<=b<=n, a!=b, 0 <= t <= 100)
The last line contains N integers, which Ai indicating the number of treasure in the ith room. (0 <= Ai <= 100)

 

 

Output

For each test case, output an integer indicating the maximum number of treasures I can take out in T minutes; if I cannot get out of the tomb, please output "Human beings die in pursuit of wealth, and birds die in pursuit of food!".

 

 

Sample Input

5 10 1 2 2 2 3 2 2 5 3 3 4 3 1 2 3 4 5

 

 

Sample Output

11

 

 

Source

2012 ACM/ICPC Asia Regional Changchun Online

 

 

Recommend

liuyiding

 

比较典型的树形DP题。比赛时没有做出来，还要加强下树形DP；

思路就是先找出1到N的路径，然后将这些路径的时间置0，之后进行树形DP

/*

树形DP

先找出1到N的边，将这些边的时间修改为0.然后就是简单的树形DP了

*/

#include<stdio.h>

#include<iostream>

#include<string.h>

#include<algorithm>

using namespace std;

const int MAXN=110;



struct Node

{

    int next;

    int to;

    int val;

}edge[MAXN*2];

int tol;

int head[MAXN];



int dp[MAXN][550];//dp[i][j]表示从第i个点开始，回到i点，花费j时间得到的最大财富值



int value[MAXN];//每个点的财富值



int time1;//从1到N需要的时间

int n;



void init()

{

    tol=0;

    memset(head,-1,sizeof(head));

}



void add(int a,int b,int val)

{

    edge[tol].to=b;

    edge[tol].next=head[a];

    edge[tol].val=val;

    head[a]=tol++;

    edge[tol].to=a;

    edge[tol].next=head[b];

    edge[tol].val=val;

    head[b]=tol++;

}



bool dfs1(int u,int pre)

{

    if(u==n)return true;//找到了

    for(int i=head[u];i!=-1;i=edge[i].next)

    {

        int v=edge[i].to;

        if(v==pre)continue;

        if(dfs1(v,u))

        {

            time1+=edge[i].val;

            edge[i].val=0;

            return true;

        }

    }

    return false;

}

int t;

void dfs2(int u,int pre)

{

    for(int i=0;i<=t;i++) dp[u][i]=value[u];

    for(int i=head[u];i!=-1;i=edge[i].next)

    {

        int v=edge[i].to;

        if(v==pre)continue;

        dfs2(v,u);

        int cost=edge[i].val*2;//要走两遍

        for(int i=t;i>=cost;i--)

           for(int j=0;j<=i-cost;j++)

             dp[u][i]=max(dp[u][i],dp[v][j]+dp[u][i-j-cost]);

    }

}

int main()

{

   // freopen("in.txt","r",stdin);

   // freopen("out.txt","w",stdout);

    int u,v,w;

    while(scanf("%d%d",&n,&t)!=EOF)

    {

        init();

        for(int i=1;i<n;i++)

        {

            scanf("%d%d%d",&u,&v,&w);

            add(u,v,w);

        }

        for(int i=1;i<=n;i++)scanf("%d",&value[i]);

        time1=0;

        dfs1(1,-1);//找从1到N的最短时间

        if(t<time1)

        {

            printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");

            continue;

        }

        t-=time1;

        dfs2(1,-1);

        printf("%d\n",dp[1][t]);

    }

    return 0;

}