hdu 1518(dfs)，hdu1518dfs
題目鏈接 ：http://acm.hdu.edu.cn/showproblem.php?pid=1518
Square
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
題意 ：給你n條邊，讓你用這些邊組成正方形（不多不少僅n條）。
分析 ：主要是超時問題，注意優化。
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1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 using namespace std;
5
6 int a[22],vis[22];
7 int n,m,length; //length表示要組成的正方形的邊長
8 int ans,flag;
9
10 void dfs(int cnt,int sum, int k) //cnt記錄邊數，sum記錄當前邊長，k記錄位置
11 {
12 if (cnt == 3) //如果有三條邊滿足要求，那麼第四條邊一定滿足要求
13 {
14 flag = 1;
15 return ;
16 }
17 if (sum == length) //找到滿足要求的邊，邊數加一，初始化
18 {
19 cnt++;
20 k = 0;
21 sum = 0;
22 }
23 for (int i=k; i<m; i++)
24 {
25 if (!vis[i] && sum + a[i] <= length)
26 {
27 vis[i] = 1;
28 dfs(cnt, sum+a[i], i+1);
29 if (flag) //優化時間，（當找到所有邊之後就一直返回，不需要再把之後的代碼運行一遍）
30 {
31 return ;
32 }
33 vis[i] = 0;
34 }
35 }
36 }
37
38 int main ()
39 {
40 scanf ("%d",&n);
41 while (n--)
42 {
43 ans = 0;
44 scanf ("%d",&m);
45 for (int i=0; i<m; i++)
46 {
47 scanf ("%d",&a[i]);
48 ans += a[i];
49 }
50 if (ans % 4) //如果所有邊長和不是4的倍數，怎樣都不能組成正方形
51 {
52 printf ("no\n");
53 continue ;
54 }
55 memset(vis, 0, sizeof(vis));
56 flag = 0;
57 length = ans / 4; //記錄正方形邊長
58 dfs(0, 0, 0);
59 if (flag)
60 printf ("yes\n");
61 else
62 printf ("no\n");
63 }
64 return 0;
65 }
View Code
