Square
Time Limit : 10000/5000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 20 Accepted Submission(s) : 12
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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
Source
University of Waterloo Local Contest 2002.09.21
此题需要优化时间，避免超时。。优化时间技巧可以学习。。。。。。。。。
1 #include <stdio.h> 2 #include<string.h> 3 int a[10000]; 4 int vist[10000]; 5 int sum; 6 int l; 7 int n; 8 int flag; 9 void Dfs(int t, int len, int index) 10 { 11 12 13 if (t == 5) 14 { 15 flag = 1; 16 return ; 17 } 18 19 if (len == l) 20 { 21 Dfs(t + 1, 0, 0); 22 if (flag)//优化时间 23 { 24 return ; 25 } 26 } 27 28 for (int i = index; i < n; i++)//从index开始优化时间 29 { 30 if (vist[i]==0 && a[i] + len <= l) 31 { 32 vist[i] = 1; 33 Dfs(t, a[i] + len, i + 1); 34 if (flag)//优化时间 35 { 36 return; 37 } 38 vist[i] = 0; 39 } 40 } 41 } 42 43 int main() 44 { 45 int i,t; 46 scanf("%d", &t); 47 while (t--) 48 { 49 50 sum = 0; 51 scanf("%d", &n); 52 for (int i = 0; i < n; i++) 53 { 54 scanf("%d", &a[i]); 55 sum += a[i]; 56 } 57 58 if (sum % 4 != 0)//简答的优化 59 { 60 puts("no"); 61 continue; 62 } 63 64 65 l = sum / 4; 66 67 68 for (i = 0; i < n; i++)//有比边长大的边就不行 69 { 70 if (a[i] > l) 71 { 72 break; 73 } 74 } 75 if (i != n) 76 { 77 puts("no"); 78 continue; 79 } 80 memset(vist, 0, sizeof(vist)); 81 flag = 0; 82 Dfs(1, 0, 0); 83 if (flag) 84 { 85 puts("yes"); 86 } 87 else 88 { 89 puts("no"); 90 } 91 } 92 return 0; 93 }