ACM水題－Rescue LK(AC，迷宮問題，DFS求解)

Rescue LK

Time Limit:1000MS  Memory Limit:65536K
Total Submit:83 Accepted:42

Description

ziliang loves LK. But LK was kidnapped by Monster.oyy,and was put in a labyrinth.
After thousands of hard fighting, ziliang finally enter the labyrinth. He must find a way to rescue LK !!! He has a map of the labyrinth, but the puzzle is too complicated, that he need your help.

Input

Input has several case. Each case first give an n and m, that is the height and length of the labyrinth. Followed by the map. In the map, ‘#’ means the wall, ‘Z’ means the position of ziliang, ‘L’
means position of LK. The ‘.’ Means empty cells.
ziliang can only walk on empty cell,and can only go up,down,left,right. And n,m will be less than 50.

Output

If ziliang can reach LK, output “LK has a good life”
Else output “Mission Failed”

Sample Input

5 5######Z..####.##..L######7 7########Z....######.######.##..#..##L.###########

Sample Output

LK has a good lifeMission Failed

Source

GDUT Monthly 2007.11 by ziliang

 

 

 

/*------------------------------------------------------------------------------------------典型的迷宮問題：通常的做法，就是用棧、DFS這兩種做法，不過歸根到底其實也是一種，都是回溯而已，棧與DFS的區別也就是遞迴與非遞迴而已。    大概想法：用DFS來進行路徑探索。每探索一步可行，就設定標誌已經走，一直走下去，直到找到終點  或者遇到不可行的格子為止。找到終點就返回1，遇到不可行就返回上一層，並重圍走過的  格子，換一個方向這樣。狀況：AC，0MS----------------------------------------------------------------------------------------*/#include<stdio.h>char chMap[52][52] ;const int nDir[4][2] = {{1,0},{0,1},{-1,0},{0,-1}} ;   //方向int m = 0 ;int n = 0 ;int zX = 0 ;//起點的X座標int zY = 0 ;//起點的Y座標int DFS(int x,int y) ;int main(void){int i = 0 ;int j = 0 ;while(scanf("%d%d",&n,&m) != EOF){getchar() ;for(i = 0 ; i < n ; ++i){for(j = 0 ; j < m ; ++j){scanf("%c",&chMap[i][j]) ;if('Z' == chMap[i][j]){zX = i ;zY = j ;}}getchar() ;}if(1 == DFS(zX,zY)){puts("LK has a good life") ;}else{puts("Mission Failed") ;}}return 0 ;}int DFS(int x,int y){int i = 0 ;int j = 0 ;if(x < 0 || y < 0 || x >= n || y >= m){return 0 ;}for(i = 0 ; i < 4 ; ++i)//換一個方向{if('L' == chMap[x+nDir[i][0]][y+nDir[i][1]]){return 1 ;}else if('.' == chMap[x+nDir[i][0]][y+nDir[i][1]]){chMap[x+nDir[i][0]][y+nDir[i][1]] = '#' ;//如果格子可行，設定已經if(1 == DFS(x+nDir[i][0],y+nDir[i][1]))//設這個方向走下去{return 1 ;}chMap[x+nDir[i][0]][y+nDir[i][1]] = '.' ;//重設原來的狀態。}}return 0 ;}