sicily 无路可逃？(图的DFS)

题意：在矩阵数组中搜索两点是否可达

解法：DFS

 1 #include<iostream>

 2 #include<memory.h>

 3 using namespace std;

 4 struct position{

 5     int x;

 6     int y;

 7     position(int xx,int yy){

 8         x = xx;

 9         y = yy;

10     }

11     position(){}

12 };

13 

14 int maze[101][101];

15 int canReach;

16 position dest;    //destination

17 int row,col;

18 int direction[4][4]={{0,1},{-1,0},{0,-1},{1,0}};

19 bool visited[101][101];

20 

21 void DFS(position p){

22     if(p.x == dest.x && p.y == dest.y){

23         canReach = 1;

24         return ;

25     }

26     for(int i=0;i<4;i++){

27         int x = p.x + direction[i][0];

28         int y = p.y + direction[i][1];

29         //check  next direction whether match the destination position or not

30         if(x == dest.x && y == dest.y){

31             canReach = 1;

32             return ;

33         }

34         //satisfy the searching condition

35         if((x <=row && x >0 )&&( y <= col && y>0) && visited[x][y] == false && maze[x][y] ==1){

36             position p1 = position(x,y);

37             visited[x][y] = true;

38             DFS(p1);

39         }

40     }

41 }

42 int main(){

43     int nCase;

44     cin >> nCase;

45     while(nCase--){    

46         int beg_x,beg_y,end_x,end_y;

47         cin >> row >> col;

48         //initialization

49         canReach = 0;

50         memset(maze,-1,sizeof(maze));

51         memset(visited,false,sizeof(visited));

52         for(int i=1;i<=row;i++)

53             for(int j=1;j<=col;j++){

54                 cin >> maze[i][j];

55             }

56         cin >> beg_x >> beg_y;

57         cin >> end_x >> end_y;

58         position p = position(beg_x,beg_y);

59         dest = position(end_x,end_y);

60         visited[beg_x][beg_y] = true;

61         //depth-first search

62         DFS(p);

63 

64         if(canReach){

65             cout<<1<<endl;

66         }else{

67             cout<<0<<endl;

68         }

69     }

70 }