USACO 4.4 Shuttle Puzzle(dfs)

比较有意思的一个题，虽然不知道为什么这么做。。。看样例就可以发现一般是w_ _b wb_ _bw只能这样4种操作。。。没想到就这样过了。。注意了一下20个换行，2Y。。。我开始还在想bfs还想过，直接构造，构造不出来。。。

/*
ID:cuizhe
LANG: C++
TASK: shuttle
*/
#include <iostream>
#include <cstdio>
#include <map>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
map<string,int> mp;
char str[32];
int que[100001];
int a[100001];
int ans = 10000000;
int n;
int judge()
{
int i;
for(i = 0; i < n; i ++)
if(str[i] != '2')
break;
if(i == n&&str
== ' ')
return 1;
else
return 0;
}
void dfs(int x,int step)
{
int i;
if(ans <= step) return;
if(judge())
{
if(step-1 < ans)
{
ans = step-1;
for(i = 1;i <= ans;i ++)
a[i] = que[i];
}
return ;
}
if(x-1 >= 0&&x-2 >= 0&&str[x-1] == '2'&&str[x-2] == '1')
{
que[step] = x-2;
str[x-2] = ' ';
str[x] = '1';
dfs(x-2,step+1);
str[x-2] = '1';
str[x] = ' ';
}
if(x-1 >= 0&&str[x-1] == '1')
{
que[step] = x-1;
str[x] = '1';
str[x-1] = ' ';
dfs(x-1,step+1);
str[x] = ' ';
str[x-1] = '1';
}
if(x+1 <= 2*n&&x+2 <= 2*n&&str[x+1] == '1'&&str[x+2] == '2')
{
que[step] = x+2;
str[x+2] = ' ';
str[x] = '2';
dfs(x+2,step+1);
str[x+2] = '2';
str[x] = ' ';
}
if(x+1 <= 2*n&&str[x+1] == '2')
{
que[step] = x+1;
str[x] = '2';
str[x+1] = ' ';
dfs(x+1,step+1);
str[x] = ' ';
str[x+1] = '2';
}
return ;
}
int main()
{
int i;
freopen("shuttle.in","r",stdin);
freopen("shuttle.out","w",stdout);
scanf("%d",&n);
for(i = 0; i < n; i ++)
{
str[i] = '1';
str[n+i+1] = '2';
}
str
= ' ';
dfs(n,1);
for(i = 1;i <= ans;i ++)
{
if(i%20 == 0||i == ans)
printf("%d\n",a[i]+1);
else
printf("%d ",a[i]+1);
}
return 0;
}