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[], n, ans;
bool vis[]; 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 ;
for(int i=pos; i<n; i++)
{
if(vis[i]) continue;
if(len+a[i]<ans)
{
vis[i] = ;
if(dfs(cur+, len+a[i], i+)) return ;
vis[i] = ;
if(len==) return ;
while(a[i]==a[i+]&&i+<n) ++i;//剪枝
}
else if(len+a[i]==ans)
{
vis[i] = ;
if(dfs(cur+, , )) return ;
vis[i] = ;
return ;
}
}
return ;
} int main()
{
int T;
scanf("%d", &T);
while(T--)
{
int Sum = ;
scanf("%d", &n);
for(int i=; i<n; i++)
{
scanf("%d", &a[i]);
Sum+=a[i];
}
if(Sum%)
{
printf("no\n");
continue;
}
ans = Sum/;
memset(vis, , sizeof(vis));
int temp = dfs(, , );
printf("%s\n", temp?"yes":"no");
}
return ;
}