hdu1518 Square（dfs）

Square

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5987    Accepted Submission(s): 1916

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

3

4 1 1 1 1

5 10 20 30 40 50

8 1 7 2 6 4 4 3 5

 

 

Sample Output

yes

no

yes

 

深搜，注意优化，就可以了

题意是：给你M根木棍，看看能不能拼出正方形来；

#include<iostream>

#include<cstring>

using namespace std;

int visit[25][2];

int len,sum,m,count,num,k,flag;

void dfs(int now,int len,int index)//now:第几条边  len:该边现在组成的长度 index:用来优化时间

{

    if (now==5)

    {

        flag=1;

        return ;

    }

    if (len==num)

    {

        dfs(now+1,0,0);

        if (flag)//优化时间

            return ;

    }



    for (int i=index;i<m;i++)//从index开始优化时间

    {

        if (!visit[i][1]&&visit[i][0]+len<=num)

        {

            visit[i][1]=1;

            dfs(now,visit[i][0]+len,i+1);//“visit[i][0]+len”不可提出来，

            if (flag)//优化时间

                return;

            visit[i][1]=0;

        }

    }

}    



int main()

{

    int n,i,j;

    scanf("%d",&n);

    while(n--)

    {

        sum=0;

        count=0;

        scanf("%d",&m);

        for(i=0;i<m;i++)

        {

            scanf("%d",&visit[i][0]);

            sum+=visit[i][0];

            visit[i][1]=0;

        }

        if(sum%4!=0||m<4||m>20)

        {

            printf("no\n");

            continue;

        }

        else

        {

            k=m/4;

            num=sum/4;

            for(i=0;i<m;i++)

                if(visit[i][0]>num)

                    {printf("no\n");break;}

            flag = 0;

            dfs(1,0,0);

            if(flag)

                printf("yes\n");

            else

                printf("no\n");

           }



    }

    return 0;

}



/*

5

8 3 4 5 6 7 8 5 2

7 4 5 6 8 10 5 2

7 4 5 6 8 10 4 3

6 3 4 4 3 1 1 

20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

*/

就这样吧，我会更加努力的