Square
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11099 Accepted Submission(s): 3566
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
题意就是：看这个数组中的数字组合是否能够构成一个正方形
不能分割数字，不能重复组合
代码：
#include<string.h> #include<stdio.h> #include<algorithm> using namespace std; int n,s,su,a[1010],vis[1010],len; bool cmp(int a,int b) { return a>b;//从大到小排序 } void dfs(int a1,int a2,int a3)//(0,0,1) { if(a1==3)/**只需要筹齐3次，那么剩下的一定能够成len长度**/ { su=1; return ; } if(su==1) return ;/**优化时间**/ for(int i=a3; i<=n; i++) { /**对于那些用过的和不符合条件的，for那里可以不扫,故从a3开始**/ /**a3前面的对于最初的a2来说一定不符合**/ if(vis[i]==0) { vis[i]=1; if(a2+a[i]==len) { dfs(a1+1,0,1); /**但是换另外一条边的时候a3要改回1,因为那些未用的，对上一条边来说不符合条件的，可能符合这条边的条件**/ } else if(a2+a[i]<len) { dfs(a1,a2+a[i],i+1); /**没筹齐从i+1继续，前面的不符合**/ while(a[i]==a[i+1]) i++; //回溯后如果后面的相同那么不需要再DFS //前面的数和后面的相同就可以跳过这个数，剪枝 } vis[i]=0; } } } int main() { int i; scanf("%d",&s); while(s--) { su=0; len=0; memset(vis,0,sizeof(vis)); //memset函数在string.h头文件中 scanf("%d",&n); for(i=1; i<=n; i++) { scanf("%d",&a[i]); len=len+a[i]; } int mm=len/4; sort(a+1,a+n+1,cmp); //在algorithm头文件中 if(len%4==0&&a[1]<=mm&&n>=4) { len/=4; dfs(0,0,1); if(su==1) printf("yes\n"); else printf("no\n"); } else { printf("no\n"); } } return 0; }
