hdoj 1518 Square 【dfs+剪枝】

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9930    Accepted Submission(s): 3254

Problem Description

Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?

Input

The first line of input contains N, the number of test cases. Each test case begins with an integer 4 <= M <= 20, the number of sticks. M integers follow; each gives the length of a stick - an integer between 1 and 10,000.

Output

For each case, output a line containing "yes" if is is possible to form a square; otherwise output "no".

Sample Input

34 1 1 1 15 10 20 30 40 508 1 7 2 6 4 4 3 5

Sample Output

yesnoyes

 

dfs+剪枝。。。

 

 

#include<stdio.h>#include<string.h>#include<math.h>#include<stdlib.h>#include<queue>#include<stack>#include<algorithm>using namespace std;bool cmp(int x,int y){    return x>y;}int aver;int stick[21];int visit[21];int n;int exist;//判断是否匹配成功 void dfs(int have,int num,int pos)//分别记录已匹配长度 已经匹配组数 当前搜索序号{    int i,j;    if(num==4)     {        exist=1;        return ;    }    for(i=pos;i<n;i++)    {        if(i&&!visit[i-1]&&stick[i]==stick[i-1])//剪枝         continue;        if(visit[i])        continue;        if(have+stick[i]==aver)        {            visit[i]=1;            dfs(0,num+1,0);            visit[i]=0;        }        else if(have+stick[i]<aver)        {            visit[i]=1;            dfs(have+stick[i],num,i+1);            visit[i]=0;        }        if(have==0)//节约时间的一步,若当前匹配失败 即使已经匹配成3对 也要跳出循环         break;//没有这句话跑800多ms 有的话只用78ms     } } int main(){    int t;    int sum;    int i,j;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        sum=0;        for(i=0;i<n;i++)        {            scanf("%d",&stick[i]);            sum+=stick[i];        }        if(sum%4!=0)        {            printf("no\n");            continue;        }        aver=sum/4;        sort(stick,stick+n,cmp);        if(stick[0]>aver)        {            printf("no\n");            continue;        }        memset(visit,0,sizeof(visit));        exist=0;        dfs(0,0,0);        if(exist)        printf("yes\n");        else        printf("no\n");    }    return 0;}