【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*];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 ;
     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=;return ;}
     }
     return ;
 }

 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=;j<=v-*t[i].c;j++) if(f[x][j]!=-)
         {
             f[y][j+*t[i].c]=mymax(f[y][j+*t[i].c],f[x][j])+w[y];
         }
         ffind(y,x);
         for(int j=;j<=v;j++) if(f[y][j]!=-)
         {
             f[x][j]=mymax(f[x][j],f[y][j]);
         }
     }
 }

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

     }
     return ;
 }

[HDU 4276]

2016-10-18 10:51:20