HDU1010 Tempter of the Bone

Tempter of the Bone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 54374    Accepted Submission(s): 14637

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 5S.X...X...XD....3 4 5S.X...X....D0 0 0

 

Sample Output

NOYES

 

Author

ZHANG, Zheng

 

Source

ZJCPC2004

 

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代码１：62ＭＳ

#include<iostream>#include<string>#include<cmath>using namespace std;char  map[10][10];int sx,sy,ex,ey;int flag;int a[4][2]={-1,0,0,-1,1,0,0,1};int n,m,t;void dfs(int x,int y,int t){    int nx,ny;    int i;  if(flag!=1)  {    if(t<fabs(ex-x)+fabs(ey-y)||(t-fabs(ex-x)+fabs(ey-y))%2);    else if(t==0)    {       if(x==ex&&y==ey)       {           flag=1;       }    }    else    {        for(i=0;i<4;i++)        {            nx=x+a[i][0];ny=y+a[i][1];            if(nx>=0&&nx<n&&ny>=0&&ny<m&&(map[nx][ny]=='.'||map[nx][ny]=='D'))            {                map[nx][ny]='D';                dfs(nx,ny,t-1);                map[nx][ny]='.';            }        }    }  }}int main(){    int i,j;    string str;    while(cin>>n>>m>>t)    {        for(i=0;i<n;i++)        {            cin>>str;            for(j=0;j<m;j++)              {                 map[i][j]=str[j];                  if(map[i][j]=='S')                    sx=i,sy=j;                  if(map[i][j]=='D')                    ex=i,ey=j;             }       }           flag=0;           dfs(sx,sy,t);           if(flag)           cout<<"YES"<<endl;           else           cout<<"NO"<<endl;    }    return 0;}

代码２：140ＭＳ

#include <iostream>using namespace std;int row,column,time;int op[4][2] = {-1,0,1,0,0,-1,0,1};char grap[10][10];int start_x,start_y,end_x,end_y;  //起点，终点int count;int sign;//当前点的X,Y坐标，t为起点到当前点的时间void dfs(int x,int y,int t){    int X,Y;    int i;    if(sign)  //如果已经找到解了，都要返回    {        return ;    }    if(end_x == x && end_y == y && t == time)  //找到目标    {        sign = 1;        return ;    }    if( (time - t) % 2 != (x+y+end_x+end_y) % 2 )  //奇偶减枝法    {        return ;    }    if(abs(x-end_x) + abs(y-end_y) > time - t)   //当前点到终点的最短时间若比剩余的时间还长的话    {        return;    }    for(i=0;i<4;i++)    {        X = x + op[i][0];        Y = y + op[i][1];        if(X <= 0 || X > row || Y <= 0 || Y > column) //判断是否越界        {            continue;        }        if(grap[X][Y] != 'X')        {            grap[X][Y] = 'X';  //将走过的.变为X            dfs(X,Y,t+1); //往下搜索，t+1            grap[X][Y] = '.';  //回溯恢复X为.        }    }}int main(){    int i,j;    while(cin>>row>>column>>time,row+column+time)    {        count = 0;        sign = 0;        for(i=1;i<=row;i++)        {            for(j=1;j<=column;j++)            {                cin>>grap[i][j];                //找到起点                if(grap[i][j] == 'S')                {                    start_x = i;                    start_y = j;                    grap[i][j] = 'X'; //将起点标记为'X',不能重复走                }                //找到终点                if(grap[i][j] == 'D')                {                    end_x = i;                    end_y = j;                }                if(grap[i][j] == '.')                {                    count++;                }            }        }        //总共可以走的步数如果比给定时间还少        if(count + 1 < time)        {            cout<<"NO"<<endl;            continue;        }        //从起点开始搜索        dfs(start_x,start_y,0);        if(sign)        {            cout<<"YES"<<endl;        }        else            cout<<"NO"<<endl;    }    return 0;}