HDU 题目1010 Tempter of the Bone

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这是一道编程题，描述了一只狗在N*M的迷宫中寻找出口的问题。迷宫会在T秒后开门，狗需要在那一秒到达门，每秒只能移动一格且不能重复经过已走过的格子。输入包含迷宫布局和开门时间，输出判断狗是否能存活（能否在T秒到达门）。样例给出了两个测试用例，其中一个无法逃脱，另一个可以逃脱。

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 125699 Accepted Submission(s): 33910

Problem Description
The doggie found a bone in an ancient maze, which 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.

Input
The input consists of multiple test cases. The first 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.

Output
For each test case, print in one line “YES” if the doggie can survive, or “NO” otherwise.

Sample Input
4 4 5
S.X.
..X.
..XD
….
3 4 5
S.X.
..X.
…D
0 0 0

Sample Output
NO
YES

杭电1010这一题的题目大意是：一条狗进入一个神奇的迷宫，一秒走一个空白的地方，狗走过的地方会消失，也就是说狗不能往回走，门在T秒之后打开，且打开很短的时间，狗只有在T秒的时候到达才能离开。

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
char mapp[10][10];
int state[10][10];
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int n,m,time,num;
int xx,yy,xxx,yyy;
bool tu;
int abs(int a)
{
    return a>0?a:-a;
}
void bfs(int x,int y,int sum)
{
    if(tu)
        return ;
    if(x==xxx&&y==yyy&&sum==time)
    {
        tu=true; return ;
    }
    int summ=abs(xxx-x)+abs(yyy-y);
    if(summ>time-sum||(summ+time-sum)%2!=0)
        return ;
    int x1,y1;
    for(int i=0;i<4;i++)
    {
        x1=x+dir[i][0];
        y1=y+dir[i][1];
        if(x1>=0&&x1<n&&y1>=0&&y1<m&&state[x1][y1]==0&&mapp[x1][y1]!='X')
        {
            state[x1][y1]=1;
            bfs(x1,y1,sum+1);
            state[x1][y1]=0;
        }
    }
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&time)!=EOF)
    {
        if(n==0&&m==0&&time==0)
            break;
        tu=false;
        num=0;
        memset(mapp,0,sizeof(mapp));
        for(int i=0;i<n;i++)
        {
            scanf("%s",mapp[i]);
            for(int j=0;j<m;j++)
            {
                state[i][j]=0;
                if(mapp[i][j]=='S')
                {
                    xx=i;
                    yy=j;
                }
                if(mapp[i][j]=='D')
                {
                    xxx=i;
                    yyy=j;
                }
                if(mapp[i][j]=='X')
                    num++;
            }
        }
        if(num+time+1>n*m)
        {
            printf("NO\n"); continue;
        }
        state[xx][yy]=1;
        bfs(xx,yy,0);
        if(tu)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}