Square
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11151 Accepted Submission(s): 3588
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
题意： 给一列数， 问能否把这列数分成四组， 且每组数的和相同。
解法： 有DFS进行回溯。
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int a[25], n, ans; bool vis[25]; int dfs(int cur, int len, int pos)//cur表示用过的个数， len表示长度， pos表示位置 { if(cur==n) return 1; for(int i=pos; i<n; i++) { if(vis[i]) continue; if(len+a[i]<ans) { vis[i] = 1; if(dfs(cur+1, len+a[i], i+1)) return 1; vis[i] = 0; if(len==0) return 0; while(a[i]==a[i+1]&&i+1<n) ++i;//剪枝 } else if(len+a[i]==ans) { vis[i] = 1; if(dfs(cur+1, 0, 0)) return 1; vis[i] = 0; return 0; } } return 0; } int main() { int T; scanf("%d", &T); while(T--) { int Sum = 0; scanf("%d", &n); for(int i=0; i<n; i++) { scanf("%d", &a[i]); Sum+=a[i]; } if(Sum%4) { printf("no "); continue; } ans = Sum/4; memset(vis, 0, sizeof(vis)); int temp = dfs(0, 0, 0); printf("%s ", temp?"yes":"no"); } return 0; }