经典搜索 剪枝 hdu 1518 Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?

主要是剪枝:分为总体和枝节

总体:sum%4!=0,max<sum/4

枝节:在于dfs的参数设置和条件判断,动态的判断是否进行

参考代码(参考牛人的):

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int a[25],vist[25],m,ave,flag;
 
int cmp(const void *a,const void *b)
{
    return (int *)b-(int *)a;
}
 
int dfs(int res,int sums,int cur)
{
    int i;
    if(sums==ave)
        return 1;
 
    for(i=cur;i<m;i++)
    {
        if(a[i]==a[i-1]&&!vist[i-1])//相邻的相等的前者没有加入,则这个也不行
            continue;
 
        if(!vist[i]&&a[i]<=res)
        {
            vist[i]=1;
 
            if(a[i]==res)
            {  
                if(dfs(ave,sums-a[i],0))
                {
                //  printf("%d\n",i);
                    return 1;
                }
            }
            else if(dfs(res-a[i],sums-a[i],i))
            {
                return 1;
            }
 
            vist[i]=0;
 
            if(res==ave)//ave没变,则表示不存在
                return 0;
        }
    }
 
    return 0;
}
 
int main()
{
    int n,i,sum;
 
    scanf("%d",&n);
 
    while(n--)
    {
        scanf("%d",&m);
 
        sum=0;
 
        for(i=0;i<m;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
 
        qsort(a,m,sizeof(int),cmp);
 
        flag=0;
 
        ave=sum/4;
 
        if(a[m-1]>ave||sum%4)
        {
            printf("no\n");
            continue;
        }else
        {
            memset(vist,0,sizeof(vist));
            if(dfs(ave,sum,0))
                flag=1;
        }
 
        if(flag)
            printf("yes\n");
        else
            printf("no\n");
 
    }
 
    return 0;
}

  

posted @   shijiwomen  阅读(2904)  评论(0编辑  收藏  举报
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