Accepted Necklace

Accepted Necklace

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3136 Accepted Submission(s): 1213

Problem Description

I have N precious stones, and plan to use K of them to make a necklace for my mother, but she won’t accept a necklace which is too heavy. Given the value and the weight of each precious stone, please help me find out the most valuable necklace my mother will accept.

Input

The first line of input is the number of cases.

For each case, the first line contains two integers N (N <= 20), the total number of stones, and K (K <= N), the exact number of stones to make a necklace.

Then N lines follow, each containing two integers: a (a<=1000), representing the value of each precious stone, and b (b<=1000), its weight.

The last line of each case contains an integer W, the maximum weight my mother will accept, W <= 1000.

Output

For each case, output the highest possible value of the necklace.

Sample Input

1

2 1

1 1

1 1

3

Sample Output

1

背包问题,DP[i][m][k]表示选第i件物品时的体积为m,以选定的物品为k件,所以Dp[i][m][k]=max(Dp[i-1][m][k],Dp[i-1][m-v[i]][k-1]+w[i]);再将其转化为01背包

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int,int>p;
const int MAX =  1100;
int Dp[MAX][35];
p Th[35];
int main()
{
    int n,k,m;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&n,&k);
        for(int i=1;i<=n;i++)
        {
            scanf("%d %d",&Th[i].first,&Th[i].second);
        }
        scanf("%d",&m);
        memset(Dp,0,sizeof(Dp));
        for(int i=1;i<=n;i++)
        {
            for(int j=m;j>=Th[i].second;j--)
            {
                for(int s=1;s<=k;s++)
                {
                    Dp[j][s]=max(Dp[j-Th[i].second][s-1]+Th[i].first,Dp[j][s]);
                }
            }
        }
        printf("%d\n",Dp[m][k]);
    }
    return 0;
}