DFS POJ 2362 Square

 

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 */