HDU 2216 Game III(BFS)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2216

Game III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1029    Accepted Submission(s): 293

Problem Description

Zjt and Sara will take part in a game, named Game III. Zjt and Sara will be in a maze, and Zjt must find Sara. There are some strang rules in this maze. If Zjt move a step, Sara will move a step in opposite direction.
Now give you the map , you shold find out the minimum steps, Zjt have to move. We say Zjt meet Sara, if they are in the same position or they are adjacent .
Zjt can only move to a empty position int four diraction (up, left, right, down). At the same time, Sara will move to a position in opposite direction, if there is empty. Otherwise , she will not move to any position.
The map is a N*M two-dimensional array. The position Zjt stays now is marked Z, and the position, where Sara stays, is marked E.

>  . : empty position
>  X: the wall
>  Z: the position Zjt now stay
>  S: the position Sara now stay

Your task is to find out the minimum steps they meet each other.

 

 

Input

The input contains several test cases. Each test case starts with a line contains three number N ,M (2<= N <= 20, 2 <= M <= 20 ) indicate the size of the map. Then N lines follows, each line contains M character. A Z and a S will be in the map as the discription above.

 

 

Output

For each test case, you should print the minimum steps. “Bad Luck!” will be print, if they can't meet each other.

 

 

Sample Input

4 4 XXXX .Z.. .XS. XXXX 4 4 XXXX .Z.. .X.S XXXX 4 4 XXXX .ZX. .XS. XXXX

 

 

Sample Output

1 1 Bad Luck!

 

 

Author

zjt

 

 

Recommend

lcy

 

这里状态需要用四维数组标记，搜索每个过程Zjt可能走的位置，然后Sara走相反的位置，如果无法走，则保持原位，注意边界问题，不然可能会RE。

之前用dfs提交RE了，不知道哪里溢出了，唉，有时间的话再找下问题所在吧

贴下AC代码：

 By LFENG

#include<cstdio> #include<cstring> #include<iostream> #include<queue> #include<cstdlib> #include<queue> using namespace std; struct node {     int sx;     int sy;     int zx;     int zy;     int time; }; char g[30][30]; int dir[4][2] = {{1, 0}, {0, 1}, { -1, 0}, {0, -1}}; int fdir[4][2] = {{ -1, 0}, {0, -1}, {1, 0}, {0, 1}}; int vis[30][30][30][30]; int n, m, x, y, x2, y2, ok; void getdata() {     int i, j;     char s[30];     for(i = 0; i < n; i++)     {         scanf("%s", s);         for(j = 0; j < m; j++)         {             if(s[j] == 'S')             {                 x = i;                 y = j;             }             else if(s[j] == 'Z')             {                 x2 = i;                 y2 = j;             }             if(s[j] == 'Z' || s[j] == 'S')g[i][j] = '.';             else g[i][j] = s[j];         }     } } void bfs() {     node p, s;     int i;     queue<node>q;     p.sx = x;     p.sy = y;     p.zx = x2;     p.zy = y2;     p.time = 0;     vis[p.sx][p.sy][p.zx][p.zy] = 1;     q.push(p);     while(!q.empty())     {         p = q.front();         q.pop();         if(abs(p.sx - p.zx) + abs(p.sy - p.zy) <= 1)         {             printf("%d\n", p.time);             return ;         }         for(i = 0; i < 4; i++)         {             s.zx = p.zx + dir[i][0];             s.zy = p.zy + dir[i][1];             if(s.zx < 0 || s.zx >= n || s.zy < 0 || s.zy >= m || g[s.zx][s.zy] == 'X')continue;             s.sx = p.sx + fdir[i][0];             s.sy = p.sy + fdir[i][1];             if(s.sx < 0 || s.sx >= n || s.sy < 0 || s.sy >= m || g[s.sx][s.sy] == 'X')             {                 s.sx = p.sx;                 s.sy = p.sy;             }             if(vis[s.sx][s.sy][s.zx][s.zy])continue;             vis[s.sx][s.sy][s.zx][s.zy] = 1;             s.time = p.time + 1;             q.push(s);         }     }     printf("Bad Luck!\n"); } int main() {     node p;     while(scanf("%d %d", &n, &m) != EOF)     {         getchar();         getdata();         memset(vis, 0, sizeof(vis));         bfs();     }     return 0; }