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

 

 

 

【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：

1 #include
      
        2 #include
       
         3 #include
        
          4 #include
         
           5 #include
          
            6 using namespace std; 7 #define Maxn 110 8 #define Maxm 510 9 10 struct node11 {12     int x,y,c,next;13 }t[Maxn*2];int len;14 int first[Maxn],w[Maxn];15 16 int mymax(int x,int y) {
    return x>y?x:y;}17 18 void ins(int x,int y,int c)19 {20     t[++len].x=x;t[len].y=y;t[len].c=c;21     t[len].next=first[x];first[x]=len;22 }23 24 int n,v;25 int dis[Maxn],sum;26 27 bool dfs(int x,int fa)28 {29     if(x==n) return 1;30     for(int i=first[x];i;i=t[i].next) if(t[i].y!=fa)31     {32         int y=t[i].y;33         dis[y]=dis[x]+t[i].c;34         if(dfs(y,x)) {sum+=t[i].c;t[i].c=0;return 1;}35     }36     return 0;37 }38 39 int f[Maxn][Maxm];40 void ffind(int x,int fa)41 {42     for(int i=first[x];i;i=t[i].next) if(t[i].y!=fa)43     {44         int y=t[i].y;45         for(int j=0;j<=v-2*t[i].c;j++) if(f[x][j]!=-1)46         {47             f[y][j+2*t[i].c]=mymax(f[y][j+2*t[i].c],f[x][j])+w[y];48         }49         ffind(y,x);50         for(int j=0;j<=v;j++) if(f[y][j]!=-1)51         {52             f[x][j]=mymax(f[x][j],f[y][j]);53         }54     }55 }56 57 int main()58 {59     while(scanf("%d%d",&n,&v)!=EOF)60     {61         len=0;62         memset(first,0,sizeof(first));63         for(int i=1;i
           
            v)73 {74 printf("Human beings die in pursuit of wealth, and birds die in pursuit of food!\n");75 }76 else77 {78 memset(f,-1,sizeof(f));79 f[1][0]=w[1];80 ffind(1,0);81 int ans=0;82 for(int i=0;i<=v-sum;i++) ans=mymax(ans,f[1][i]);83 printf("%d\n",ans);84 }85 86 }87 return 0;88 }
           
          
         
        
       
      

 

2016-10-18 10:51:20