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 题意：给出宝石的数目n，制成项链所需的宝石个数k，然后再给出每个宝石的价值与重量，还有母亲会接受的最大价值，求出在最大价值范围内，项链的价值尽可能大。思路：这一次用dfs做了，等掌握了二维费用背包之后再尝试用背包去做
[cpp]
#include <stdio.h>
#include <string.h>
struct node {
int v,w; }p[50];
int maxn,n,k,weight,vis[50];
   void dfs(int wei,int val,int step,int sum)
{      if(sum == k || wei == weight)
{          if(maxn<val)          maxn = val;
        return ;
   }      for(int i = step;i<=n;i++)
     {          if(!vis[i] && sum+1<=k && wei+p[i].w<=weight)
        {              vis[i] = 1;
         dfs(wei+p[i].w,val+p[i].v,i+1,sum+1);
           vis[i] = 0;          }
}      return ; }    int
main() {      int t,i,j;
scanf("%d",&t);
while(t--)      {
   scanf("%d%d",&n,&k);
    for(i = 1;i<=n;i++)
    {              scanf("%d%d",&p[i].v,&p[i].w);
           vis[i] = 0;          }
   scanf("%d",&weight);
      maxn = 0;
       dfs(0,0,0,0);
     printf("%d\n",maxn);
}        return 0; }

补充：软件开发 , C++ ,