题目传送门
1 /*
2 DFS：问能否用小棍子组成一个正方形
3 剪枝有3：长的不灵活，先考虑；若根本构不成正方形，直接no；若第一根比边长长，no
4 这题是POJ_1011的精简版:)
5 */
6 #include <cstdio>
7 #include <iostream>
8 #include <cstring>
9 #include <map>
10 #include <set>
11 #include <cmath>
12 #include <algorithm>
13 using namespace std;
14
15 const int MAXN = 22;
16 const int INF = 0x3f3f3f3f;
17 int a[MAXN];
18 bool vis[MAXN];
19 int len, sum;
20 int n, m;
21
22 bool cmp(int x, int y)
23 {
24 return x > y;
25 }
26
27 bool DFS(int ans, int cnt, int s)
28 {
29 if (cnt == 3)
30 {
31 return true;
32 }
33
34 for (int i=s; i<=m; ++i)
35 {
36 if (!vis[i] && ans + a[i] <= len)
37 {
38 vis[i] = true;
39 if (ans + a[i] == len)
40 {
41 if (DFS (0, cnt + 1, 1) == true) return true;
42 vis[i] = false;
43 }
44 else
45 {
46 if (DFS (ans + a[i], cnt, i) == true) return true;
47 vis[i] = false;
48 }
49 }
50 }
51
52 return false;
53 }
54
55 int main(void) //POJ 2362 Square
56 {
57 //freopen ("POJ_2362.in", "r", stdin);
58
59 scanf ("%d", &n);
60 while (n--)
61 {
62 sum = 0;
63 memset (vis, 0, sizeof (vis));
64
65 scanf ("%d", &m);
66 for (int i=1; i<=m; ++i)
67 {
68 scanf ("%d", &a[i]);
69 sum += a[i];
70 }
71 sort (a+1, a+1+m, cmp); //Cut 1
72
73 if (m < 4 || sum % 4 != 0) //Cut 2
74 {
75 puts ("no"); continue;
76 }
77 len = sum / 4;
78
79 if (a[1] > len) //Cut 3
80 {
81 puts ("no"); continue;
82 }
83
84 if (DFS (0, 0, 1) == true) puts ("yes");
85 else puts ("no");
86 }
87
88 return 0;
89 }
90
91
92 /*
93 yes
94 no
95 yes
96 */