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

#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;
}