Hdu 2602 Bone Collector 01背包 解题报告

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 ?

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.

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

思路

很裸

代码

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<cmath>#include<algorithm>using namespace std;const int N=1005;int dp[N][N],val[N],cost[N],T,n,v;int main(){    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&v);        for (int i=1;i<=n;i++)        scanf("%d",&val[i]);        for (int i=1;i<=n;i++)        scanf("%d",&cost[i]);        memset(dp,0,sizeof(dp));        for (int i=1;i<=n;i++)        for (int j=0;j<=v;j++)        {            if (cost[i]<=j) dp[i][j]=max(dp[i-1][j],dp[i-1][j-cost[i]]+val[i]);            else dp[i][j]=dp[i-1][j];        }        printf("%d\n",dp[n][v]);    }    return 0;}