TJU 1398 Square DFS
**Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square?
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.
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
Output for Sample Input
yes
no
yes
**
题意:给出n条边,判断这些边是否能够组成正方形
思路:
剪枝部分:第一种情况:边长总长度必须是4的倍数
第二种情况:正方形边长必须大于最长的边
第三种情况:边的总数必须大于四条边
搜索部分:从最长的边依次往最短的边搜索,满足边长总长度为所需边长长度时才继续搜索满足下一条边的情况,用过一条边记得标记一下,如果凑齐了三条边,则必定有解。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int f[25],num[25];
int len,tn;
/*pos表示当前位置,l表示当前长度,count表示已找到的边长*/
int dfs(int pos,int l,int count)
{
int i;
/*只需要找到三条边即可*/
if( count == 3)
return 1;
for( i = pos; i >= 1; i --)
{
/*用f数组记录该边长是否被找过*/
if(!f[i])
{
f[i] = 1;
/*如果没有达到边长的长度*/
if( l+num[i] < len)
{
if(dfs(i-1,l+num[i],count))
return 1;
}
/*如果满足边长的长度*/
else if( l+num[i] == len)
{
/*继续从最长边开始找,已找到的边长总数得加1*/
if( dfs(tn,0,count+1))
return 1;
}
f[i] = 0;/*回溯,置为初始状态,便于下次搜索*/
}
}
return 0;
}
int main()
{
int t,i,sum;
scanf("%d",&t);
while( t--)
{
scanf("%d",&tn);
sum = 0;
memset(f,0,sizeof(f));
for( i = 1; i <= tn; i ++)
{
scanf("%d",&num[i]);
sum += num[i];
}
sort(num+1,num+tn+1);
len = sum/4;
/*剪枝,判断三种情况*/
if( len*4!=sum||len < num[tn]||tn < 4)
{
printf("no\n");
continue;
}
/*从最长的边开始搜索*/
if( dfs(tn,0,0))
{
printf("yes\n");
}
else
printf("no\n");
}
return 0;
}
后记:这道题,哎,真是,错了9遍才过,剪枝漏掉了边长总数小于4的情况,搜索部分错在凑齐该条边以后应该重新从最长的边搜索。