Accepted Necklace hdu 2660

Accepted Necklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1022    Accepted Submission(s): 406

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

 

Source

HDU男生专场公开赛——赶在女生之前先过节（From WHU）

 

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#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int T, N, K, W;
struct node
{
    int v, w;
}P1[100];

int sum = 0;

void DFS(int v, int w, int num, int j )
{
    
      int i;
    
    if ( w > W)
        return ;
    if (num > K)
        return;
    if (sum < v)
        sum = v;
    for (i = j; i < N; i++) {
        DFS( v + P1[i].v,  w + P1[i].w, num + 1, i + 1);
    }
    
}

int main( )
{
    int i, j;
    
    scanf("%d", &T);
    while( T--)
    {   
        sum = 0;
        scanf("%d%d", &N, &K);
        for(i = 0; i < N; i++)
            scanf("%d%d", &P1[i].v, &P1[i].w);
        scanf("%d",&W);
        DFS(0, 0, 0, 0);// 参数传递，  当前价值，当值重量， 当前个数 , 数组下标 
        printf("%d\n",sum);
    }
    return 0;
}