hdu2010(dfs+剪枝)
Tempter of the Bone
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 131619 Accepted Submission(s):
35432
fascinated him a lot. However, when he picked it up, the maze began to shake,
and the doggie could feel the ground sinking. He realized that the bone was a
trap, and he tried desperately to get out of this maze.
The maze was a
rectangle with sizes N by M. There was a door in the maze. At the beginning, the
door was closed and it would open at the T-th second for a short period of time
(less than 1 second). Therefore the doggie had to arrive at the door on exactly
the T-th second. In every second, he could move one block to one of the upper,
lower, left and right neighboring blocks. Once he entered a block, the ground of
this block would start to sink and disappear in the next second. He could not
stay at one block for more than one second, nor could he move into a visited
block. Can the poor doggie survive? Please help him.
line of each test case contains three integers N, M, and T (1 < N, M < 7;
0 < T < 50), which denote the sizes of the maze and the time at which the
door will open, respectively. The next N lines give the maze layout, with each
line containing M characters. A character is one of the following:
'X': a
block of wall, which the doggie cannot enter;
'S': the start point of the
doggie;
'D': the Door; or
'.': an empty block.
The input is
terminated with three 0's. This test case is not to be processed.
doggie can survive, or "NO" otherwise.
..X.
..XD
....
3 4 5
S.X.
..X.
...D
0 0 0
#include<iostream>
#include<string.h>
using namespace std;
#include<math.h>
char s[][];
int ax,ay,bx,by,n,m,k;
int t[][]={,,-,,,,,-},visit[][],flag;
void dfs(int x,int y,int count)
{
int mx,my,i;
if(x==bx&&y==by)
{
if(k==count){ flag=;
}
return;
}
if(count>=k)
return;
if(s[x][y]!='X')
{
for(i=;i<;i++)
{
mx=x+t[i][];
my=y+t[i][];
if(s[mx][my]!='X'&&mx>=&&mx<=n&&my>=&&my<=m&&!visit[mx][my])
{
visit[mx][my]=;
dfs(mx,my,count+);
visit[mx][my]=;
if(flag) //注意,在找到了目标之后,就不需要再找!以往编写dfs时,没有注意这点,就会超时
return;
}
}
}
}
int main()
{
while(cin>>n>>m>>k)
{
if(n==&&m==&&k==)
return ;
int i,count=;
flag=;
for(i=;i<=n;i++)
{
getchar();
for(int j=;j<=m;j++)
{
cin>>s[i][j];
if(s[i][j]=='S')
{
ax=i;ay=j;
}
if(s[i][j]=='D')
{
bx=i;by=j;
} }
}
getchar();
memset(visit,,sizeof(visit));
if((abs(ax-bx)+abs(ay-by))>k||(ax+ay+bx+by+k)%==)//剪枝干
{
//cout<<"*"<<endl;
cout<<"NO"<<endl;
continue;
}
visit[ax][ay]=;
count=;
dfs(ax,ay,count);
if(flag==)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return ;
}
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