Square
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17297 Accepted Submission(s):
5419
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
//剪枝很重要 #include <iostream> #include <algorithm> using namespace std; long long a[25],n,s; bool v[25]; int m; bool cmp(long long x,long long y) { return x>y; } int dfs(long long h, int w, int k) { if(h == s) { w = 1; h = 0; k++; if(k == 4) return 1; } for(int i=w; i<=m; i++) { if(!v[i]) { v[i] = 1; if((h+a[i] <= s ) && dfs(h+a[i], i+1, k)) //有一根无法和其他的拼就直接跳出来 //而不是一根根选 return 1; v[i] = 0; } } return 0; } int main() { scanf("%lld",&n); while(n--) { scanf("%d",&m); s=0; int i; for(i=1;i<=m;i++) { scanf("%lld",&a[i]); v[i]=0; s=s+a[i]; } if(s%4!=0) {printf("no ");continue;} s=s/4; for(i=1;i<=m;i++) { if(a[i]>s) {printf("no ");break;} } if(i!=m+1) continue; sort(a+1,a+m+1,cmp); if(a[0]+a[m-1]>s) {printf("no ");break;} if(dfs(0,1,0)) printf("yes "); else printf("no "); } return 0; }