hdu 2602(dp)

2019-11-17 02:42:03

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hdu 2602(dp)

题目链接：http://acm.hdu.edu.cn/showPRoblem.php?pid=2602

Bone Collector

Time Limit: 2000/1000 MS (java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 32499Accepted Submission(s): 13379

Problem DescriptionMany 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 ?

InputThe 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.

OutputOne integer per line representing the maximum of the total value (this number will be less than 231).

Sample Input1 5 10 1 2 3 4 5 5 4 3 2 1

Sample Output14题意：给你一个包的体积，每块骨头的价值和占用的体积，求出可以放入价值最大方案的价值。分析：简单的01背包，纯属模板题，也是我做的第一题背包，就直接贴代码了。

 1 #include <cstdio> 2 #include <cmath> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6  7 //dp[i][j]表示放入第i块骨头并占用j体积的最大价值， 8 //c[i]表示第i块骨头的体积 9 //w[i]表示第i块骨头的价值10 11 int dp[1111][1111],c[1111],w[1111];12 int T,N,V;13 14 int main ()15 {16     int i,j;17     scanf ("%d",&T);18     while (T--)19     {20         scanf ("%d%d",&N,&V);21         for (i=1; i<=N; i++)22         scanf ("%d",&w[i]);23         for (i=1; i<=N; i++)24         scanf ("%d",&c[i]);25         memset(dp, 0, sizeof(dp));26         for (i=1; i<=N; i++)27         {28             for (j=0; j<=V; j++)29             {30                 dp[i][j] = dp[i-1][j];   //这里主要考虑j小于c[i]时放不下第i块骨头31                 if (j >= c[i])32                 dp[i][j] = max(dp[i-1][j], dp[i-1][j-c[i]]+w[i]);33             }34         }35         printf ("%d/n",dp[N][V]);36     }37     return 0;38 }

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