更深层次的DFS 解决POJ 2362

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

//qsort的用法 //qsort(数组名，数组大小，数组的类型，比较的函数) //比较的函数的定义，见19行

View Code

#include <stdio.h>

#include <stdlib.h>

#include <string.h>



int part, N;

int len[128];

char used[128];



int cmp(const void *a, const void *b)  //qsort的自定义函数

{

    return *(int *)b - *(int *)a;

}



int dfs(int side, int start, int sidelength)

{

    int i;



    if (sidelength==0) 

    {

        sidelength = part;  //再次赋值为边长的大小

        side++;  //用来统计满足条件的边的个数

        start = 0;

    }



    if (side == 4)  //函数的出口，当有4个边时，正好构成正方形，返回真,作整个DFS的出口

        return 1;



    for (i = start; i < N; i++) 

    {

        if (len[i] > sidelength || used[i] )  //大于边长 或者 已经被用过

            continue;



        used[i] = 1;  //否则将其标记为已经用过

        if (dfs(side, i + 1, sidelength - len[i]))

        {

            return 1;  //只有当30行的return 1执行之后,才会执行这里的return 1;结束DFS

        }

        used[i] = 0;

    }

    return 0;

}



int solve()

{

    int i, sum;



    scanf("%d", &N);

    sum = 0;

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

    {

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

        sum += len[i];

    }



    // 1

    if (sum % 4)

        return 0;



    part = sum / 4;

    qsort(len, N, sizeof(len[0]), cmp); //数组名，数组大小，数组的类型，比较的函数

    // 2

    if (len[0] > part)

        return 0;



    memset(used, 0, N);

    return dfs(0, 0, part);

}



int main()

{

    int t;



    scanf("%d", &t);

    while (t--) 

        printf("%s\n", solve() ? "yes" : "no");



    return 0;

}