hdu- 2602 Bone Collector
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 2
31).
这道题目是我做的第一道背包问题,以前没注意过,以后会多学这方面的东西。
核心代码:
dp[j]=max(dp[j],dp[j-c[i]]+w[i]);
所有代码:
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[1200];
int w[1200];
int c[1200];
int main(){
int T;
int i,j,k,t;
int n,v;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&v);
for(i=0;i<n;i++)
scanf("%d",&w[i]);
for(i=0;i<n;i++)
scanf("%d",&c[i]);
memset(dp,0,sizeof(dp));
for(i=0;i<n;i++){
for(j=v;j>=c[i];j--)
dp[j]=max(dp[j],dp[j-c[i]]+w[i]);
}
printf("%d\n",dp[v]);
}
return 0;
}
Sample Input
1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output
14

浙公网安备 33010602011771号