hdu 2660 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 <= ), 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<=), representing the value of each precious stone, and b (b<=), its weight.
The last line of each case contains an integer W, the maximum weight my mother will accept, W <= . 

Output

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

 

Sample Input

Sample Output

 

Source

HDU男生专场公开赛——赶在女生之前先过节（From WHU）

 

dfs暴搜，不断剪枝就过了，不过在

for(int i=cur;i<n;i++){
if(!vis[i]){
vis[i]=1;
dfs(i,num+1,weight+w[i],value+v[i]);
vis[i]=0;
}
}

这里一开写的是for(int i=cur+1;i<n;i++)就错了，改成for(int i=cur;i<n;i++)就莫名其妙地A了。

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 #include <stack>
 using namespace std;
 #define PI acos(-1.0)
 #define max(a,b) (a) > (b) ? (a) : (b)
 #define min(a,b) (a) < (b) ? (a) : (b)
 #define ll long long
 #define eps 1e-10
 #define MOD 1000000007
 #define N 26
 #define inf 1<<26
 int n,k,total;
 int ans;
 int v[N];
 int w[N];
 int vis[N];
 void dfs(int cur,int num,int weight,int value){
    if(weight>total){
       return;
    }
    int sum=;
    for(int i=cur+;i<n;i++){
       sum+=v[i];
    }
    if(value+sum<ans){
       return;
    }
    if(num==k){
       ans=max(ans,value);
       return;
    }
    for(int i=cur;i<n;i++){
       if(!vis[i]){
          vis[i]=;
          dfs(i,num+,weight+w[i],value+v[i]);
          vis[i]=;
       }
    }
 }
 int main()
 {
    int t;
    scanf("%d",&t);
    while(t--){
       scanf("%d%d",&n,&k);
       for(int i=;i<n;i++){
          scanf("%d%d",&v[i],&w[i]);
       }
       scanf("%d",&total);
       memset(vis,,sizeof(vis));
       ans=-inf;
       dfs(,,,);
       printf("%d\n",ans);
    }
     return ;
 }