【HDU 4276】The Ghost Blows Light（树形DP，依赖背包）

The Ghost Blows Light

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

 

【题意】

　　一个有 N 个节点的树形的地图，知道了每条边经过所需要的时间，现在给出时间T，问能不能在T时间内从 1号节点到 N 节点。每个节点都有相对应的价值，而且每个价值只能被取一次，问如果可以从1 号节点走到 n 号节点的话，最多可以取到的最大价值为多少。

【分析】

　　这题跟poj2486类似，不过这题的问题是不知道是否回到n。
　　规定回到n的话，我们画一下图会发现，我们走的路径是1到n的路径只走一遍，其他路径一定是去和回的两遍。所以我们可以先求出1~n的路径，然后把这些边权置为0，然后就是依赖背包问题，f[i][j]表示在i这棵子树上走j分钟的最大价值，我们走一条边的费用是*2的，因去和回是两遍，最后要加上1~n的路径的代价。
　　（网上打的都是树形0-1背包是n^3，依赖背包可以打成n^2）

代码如下：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define Maxn 110
#define Maxm 510

struct node
{
    int x,y,c,next;
}t[Maxn*2];int len;
int first[Maxn],w[Maxn];

int mymax(int x,int y) {return x>y?x:y;}

void ins(int x,int y,int c)
{
    t[++len].x=x;t[len].y=y;t[len].c=c;
    t[len].next=first[x];first[x]=len;
}

int n,v;
int dis[Maxn],sum;

bool dfs(int x,int fa)
{
    if(x==n) return 1;
    for(int i=first[x];i;i=t[i].next) if(t[i].y!=fa)
    {
        int y=t[i].y;
        dis[y]=dis[x]+t[i].c;
        if(dfs(y,x)) {sum+=t[i].c;t[i].c=0;return 1;}
    }
    return 0;
}

int f[Maxn][Maxm];
void ffind(int x,int fa)
{
    for(int i=first[x];i;i=t[i].next) if(t[i].y!=fa)
    {
        int y=t[i].y;
        for(int j=0;j<=v-2*t[i].c;j++) if(f[x][j]!=-1)
        {
            f[y][j+2*t[i].c]=mymax(f[y][j+2*t[i].c],f[x][j])+w[y];
        }
        ffind(y,x);
        for(int j=0;j<=v;j++) if(f[y][j]!=-1)
        {
            f[x][j]=mymax(f[x][j],f[y][j]);
        }
    }
}

int main()
{
    while(scanf("%d%d",&n,&v)!=EOF)
    {
        len=0;
        memset(first,0,sizeof(first));
        for(int i=1;i<n;i++)
        {
            int x,y,c;
            scanf("%d%d%d",&x,&y,&c);
            ins(x,y,c);ins(y,x,c);
        }
        for(int i=1;i<=n;i++) scanf("%d",&w[i]);
        sum=0;
        dfs(1,0);
        if(sum>v)
        {
            printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!
");
        }
        else
        {
            memset(f,-1,sizeof(f));
            f[1][0]=w[1];
            ffind(1,0);
            int ans=0;
            for(int i=0;i<=v-sum;i++) ans=mymax(ans,f[1][i]);
            printf("%d
",ans);
        }
        
    }
    return 0;
}

2016-10-18 10:51:20