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

 #include<iostream>
 #include<memory.h>
 using namespace std;
 struct position{
     int x;
     int y;
     position(int xx,int yy){
         x = xx;
         y = yy;
     }
     position(){}
 };

 int maze[][];
 int canReach;
 position dest;    //destination
 int row,col;
 int direction[][]={{,},{-,},{,-},{,}};
 bool visited[][];

 void DFS(position p){
     if(p.x == dest.x && p.y == dest.y){
         canReach = ;
         return ;
     }
     for(int i=;i<;i++){
         int x = p.x + direction[i][];
         int y = p.y + direction[i][];
         //check  next direction whether match the destination position or not
         if(x == dest.x && y == dest.y){
             canReach = ;
             return ;
         }
         //satisfy the searching condition
         if((x <=row && x > )&&( y <= col && y>) && visited[x][y] == false && maze[x][y] ==){
             position p1 = position(x,y);
             visited[x][y] = true;
             DFS(p1);
         }
     }
 }
 int main(){
     int nCase;
     cin >> nCase;
     while(nCase--){
         int beg_x,beg_y,end_x,end_y;
         cin >> row >> col;
         //initialization
         canReach = ;
         memset(maze,-,sizeof(maze));
         memset(visited,false,sizeof(visited));
         for(int i=;i<=row;i++)
             for(int j=;j<=col;j++){
                 cin >> maze[i][j];
             }
         cin >> beg_x >> beg_y;
         cin >> end_x >> end_y;
         position p = position(beg_x,beg_y);
         dest = position(end_x,end_y);
         visited[beg_x][beg_y] = true;
         //depth-first search
         DFS(p);

         if(canReach){
             cout<<<<endl;
         }else{
             cout<<<<endl;
         }
     }
 }