HDOJ Accepted Necklace

最新推荐文章于 2020-01-13 16:30:40 发布

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最新推荐文章于 2020-01-13 16:30:40 发布

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分类专栏： DFS and BFS 文章标签： each output integer input

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Accepted Necklace

http://acm.hdu.edu.cn/diy/contest_showproblem.php?cid=12573&pid=1006

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 11 Accepted Submission(s) : 4

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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

Sample Output

Source

HDU男生专场公开赛——赶在女生之前先过节（From WHU）

这题主要考虑当前状态，否则会超时的。

#include<iostream>
using namespace std;
int visited[25],weight[25],value[25];
int N,K,W;
int maxval; //最优解
void dfs(int p,int num,int val,int wei ){ //p表示当前状态，初始化是1
     if(num==K)
        if(val>maxval)
           maxval=val;
     for(int i=p;i<=N;i++){
         if(!visited[i]){
              if((num+1)<=K && (wei+weight[i])<=W ){ //这一步剪枝
                  visited[i]=1;
                  dfs(i+1,num+1,val+value[i],wei+weight[i]);
                  visited[i]=0;
              }
         }
     }
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
      maxval=0;
      scanf("%d%d",&N,&K);
      for(int i=1;i<=N;i++)
          visited[i]=0,scanf("%d%d",&value[i],&weight[i]);
      scanf("%d",&W);
      dfs(1,0,0,0);
      printf("%d\n",maxval);
    }
    return 0;
}