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 }
