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.

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.

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

1
2 1
1 1
1 1
3 

先超时杭电上提交，后加了个判重，修改前：

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;int val[1002],wei[1002];
int sum;
int case1,n,k,w;
int best;
bool vis[1002];void dfs(int v,int we,int sum)
{
    int i;
    if(sum>k)        return ;
    if(we>w)         return ;
    if(sum==k&&we<=w&&best<v)  best=v;
    for(i=1;i<=n;i++)
    {
        if(!vis[i]&&we<w&&sum<k)
        {
            vis[i]=1;
             v+=val[i];
             we+=wei[i];
             sum++;
             dfs(v,we,sum);
             v-=val[i];
             we-=wei[i];
             sum--;
             vis[i]=0;
        }
    }
}int main()
{
    int i;scanf("%d",&case1);
    while(case1--)
    {
        cin>>n>>k;
        for(i=1;i<=n;i++)
        {
         cin>>val[i]>>wei[i];
        }
        cin>>w;
        memset(vis,0,sizeof(vis));
        best=0;
        dfs(0,0,0);
        printf("%d\n",best);
    }
    return 0;
}

修改后：

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;int val[1002],wei[1002];
int sum;
int case1,n,k,w;
int best;
bool vis[1002];void dfs(int v,int we,int sum,int first)
{
    int i;
    if(sum>k)        return ;
    if(we>w)         return ;
    if(sum==k&&we<=w&&best<v)  best=v;
    for(i=first;i<=n;i++)
    {
        if(!vis[i]&&we<w&&sum<k)
        {
            vis[i]=1;
             v+=val[i];
             we+=wei[i];
             sum++;
             dfs(v,we,sum,i+1);
             v-=val[i];
             we-=wei[i];
             sum--;
             vis[i]=0;
        }
    }
}int main()
{
    int i;scanf("%d",&case1);
    while(case1--)
    {
        cin>>n>>k;
        for(i=1;i<=n;i++)
        {
         cin>>val[i]>>wei[i];
        }
        cin>>w;
        memset(vis,0,sizeof(vis));
        best=0;
        dfs(0,0,0,1);
        printf("%d\n",best);
    }
    return 0;
}

参考网上的代码：

这道题目是DFS题目，但又参杂了多重背包问题。开始我了3次都是Time limited。。

后来才知道又有多重背包问题，多重背包即要考虑重复问题，

即加个参数考虑下重复的情况就O了AC啦。

关键点： DFS考虑第1是否用时，然后递归用后面的2--N个，组合起来看是否组成最大价值的

项链，将其保存；接下来循环就从第2个开始，不需要在管第1个背包了，否则重复；

举例： 3 4 5 6 == 4 3 5 6 == 5 3 6 4这个道理，所以下面代码中的DFS中FOR循环是

START开始的。