hdu2602 简单01背包
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Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 52591 Accepted Submission(s): 22165
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 ?
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.
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 231).
Sample Input
15 101 2 3 4 55 4 3 2 1
Sample Output
14
简单01背包
#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int value[1050];int f[1050];int dp[1050];int main(){ int T; scanf("%d",&T); while(T--){ memset(dp,0,sizeof(dp)); int n,total; scanf("%d%d",&n,&total); for(int i = 1; i <= n; i++) scanf("%d",&value[i]); for(int i = 1; i <= n; i++) scanf("%d",&f[i]); for(int i = 1; i <= n; i++)//!!n个骨头 for(int j = total; j >= f[i]; j--) dp[j] = max(dp[j],dp[j- f[i]]+value[i]); cout<< dp[total] << endl; }}
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