Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 29109 Accepted Submission(s): 11898

Problem Description Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.
Output One integer per line representing the maximum of the total value (this number will be less than 2 ³¹).
Sample Input

Sample Output

Author Teddy
Source HDU 1st “Vegetable-Birds Cup” Programming Open Contest 这是 01 背包问题 01 背包问题用到了递推的思想，并且用到了分治将大问题转化为能够容易解决的，小问题，从小问题来得到答案的思想大致思路如下： c[i-1][v]表示前i-1件物品恰放入一个质量为m的背包可以取到最大价值，若只考虑第i件物品的话，那么有两种情况，就是放，与不放，要是放的话，那么问题就转化为“将前n-1件物品放入剩下的容量为m-w[i]的背包中，此时获得的最大价值就是c[i-1][m-w[i]]再加上最后放进去的第i件的物品所获得价值p[i];要是不放的话，问题就变成：”将前n-1件物品放入容量为j的背包中，此时获得的最大价值就是c[i-1][m];然后题上一般有物品的数量，及相应的价值，通过两个for循环就好了；最后的最大价值就是c[n][m]; 代码如下：

#include
  
   
#include
   
     int dp[1010][1010],val[1010],vol[1010]; int main() { int n,i,j,v,t; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&v); for(i=1;i<=n;i++) scanf("%d",&val[i]); for(j=1;j<=n;j++) scanf("%d",&vol[j]); memset(dp,0,sizeof(dp)); for(i=1;i<=n;i++) for(j=0;j<=v;j++)//题上有个陷阱，就是将没有体积但是有价值的 骨头考虑了进去 { if(vol[i]<=j&&dp[i-1][j]


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