hdu 2602 - Bone Collector(01背包)解题报告
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Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 39532 Accepted Submission(s): 16385
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
题解:
裸的01背包,经典例题。
参考代码:
#include<stdio.h>#define M 1111#include<string.h>int max(int a,int b){return a>b?a:b;}int main(){int t,a[M],b[M],dp[M];scanf("%d",&t);while(t--){int n,v;memset(dp,0,sizeof(dp));scanf("%d%d",&n,&v);for(int i=0;i<n;i++) scanf("%d",&b[i]); for(int i=0;i<n;i++) scanf("%d",&a[i]); for(int i=0;i<n;i++) { for(int j=v;j>=a[i];j--) { dp[j]=max(dp[j],dp[j-a[i]]+b[i]); } } printf("%d\n",dp[v]);}return 0;}
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