Square
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6187 Accepted Submission(s): 2013
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
Source
Recommend
LL
思路：DFS
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int map[100];
int n,m;
int used[100];
int sum;
int side;
int flag;
int mmm;
void DFS(int now,int cnt,int k)
{
if(now == side)
{
cnt ++;
now = 0;
k = 0;
}
if(cnt == 4)
{
flag = 1;
return ;
}
for(int i = k;i <= m;i ++)
{
if(!used[i])
{
used[i] = 1;
if(now + map[i] <= side)
{
DFS(now + map[i],cnt,i + 1);
if(flag)
return ;
}
used[i] = 0;
}
}
}
int main()
{
scanf("%d",&n);
while(n --)
{
scanf("%d",&m);
sum = 0;
mmm = 0;
flag = 0;
for(int i = 1;i <= m;i ++)
{
scanf("%d",&map[i]);
if(mmm < map[i])
mmm = map[i];
sum += map[i];
}
if(sum % 4 != 0 || mmm > sum / 4)
{
flag = 0;
}
else
{
side = sum / 4;
memset(used,0,sizeof(used));
DFS(0,0,0);
}
if(flag == 1)
printf("yes ");
else
printf("no ");
}
return 0;
}
#include <cstdio>
#include <cstring>
using namespace std;
int map[100];
int n,m;
int used[100];
int sum;
int side;
int flag;
int mmm;
void DFS(int now,int cnt,int k)
{
if(now == side)
{
cnt ++;
now = 0;
k = 0;
}
if(cnt == 4)
{
flag = 1;
return ;
}
for(int i = k;i <= m;i ++)
{
if(!used[i])
{
used[i] = 1;
if(now + map[i] <= side)
{
DFS(now + map[i],cnt,i + 1);
if(flag)
return ;
}
used[i] = 0;
}
}
}
int main()
{
scanf("%d",&n);
while(n --)
{
scanf("%d",&m);
sum = 0;
mmm = 0;
flag = 0;
for(int i = 1;i <= m;i ++)
{
scanf("%d",&map[i]);
if(mmm < map[i])
mmm = map[i];
sum += map[i];
}
if(sum % 4 != 0 || mmm > sum / 4)
{
flag = 0;
}
else
{
side = sum / 4;
memset(used,0,sizeof(used));
DFS(0,0,0);
}
if(flag == 1)
printf("yes ");
else
printf("no ");
}
return 0;
}