hdu2660 Accepted Necklace (DFS)

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

 

题目意思：求出K个宝石最大价值总和，但重量不能超过W；

#include<stdio.h>

struct ston

{

    int sa,sw;

};

struct ston s[25],tem;

int su,N,K,W;

void DFS(int i,int suma, int w,int k)

{

    int j;

    if(su<suma)//比较总价值

        su=suma;

    if(k==K)//宝石个数不能超过K个

    return ;

    for(j=i+1;j<=N;j++)

    if(s[j].sw+w<=W&&k+1<=K)

    DFS(j,s[j].sa+suma,s[j].sw+w,k+1);

}

int main()

{

    int t,i,j,e,sum;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d%d",&N,&K);

        for(i=1;i<=N;i++)

        scanf("%d%d",&s[i].sa,&s[i].sw);

        scanf("%d",&W);



        for(i=1;i<=N;i++)//先按价值从大到小排序

        {

            e=i;

            for(j=i+1;j<=N;j++)

            if(s[e].sa<s[j].sa)

            e=j;

            tem=s[i];s[i]=s[e];s[e]=tem;

        }

        sum=0;

        for(i=1;i<=N;i++)//看以那个开头总价值最大

        if(s[i].sw<=W&&K>0)

        {

            su=s[i].sa;

            DFS(i,s[i].sa,s[i].sw,1);

            if(su>sum)

            sum=su;

        }

        printf("%d\n",sum);

    }

}