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 ?
![](http://acm.hdu.edu.cn/data/images/C154-1003-1.jpg)
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 ?
![](http://acm.hdu.edu.cn/data/images/C154-1003-1.jpg)
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
1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output
14
题解:
裸的01背包,经典例题。
參考代码:
#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; }