There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.
Write a program to count the number of black tiles which he can reach by repeating the moves described above.InputThe input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.
There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.
'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)
OutputFor each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).
Sample Input6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0Sample Output
45 59 6 13
题目大意：“#”相当于不能走的陷阱或墙壁，“.”是可以走的路。从@点出发，统计所能到达的地点总数
1 //并查集解决 2 #include<iostream> 3 using namespace std; 4 5 const int h = 22; 6 char map[h][h]; 7 int key[h*h]; 8 int rrank[h*h]; 9 int n,m,dx,dy; 10 11 int find(int a){ 12 return a==key[a]? a : key[a]=find(key[a]); 13 } 14 15 void key_union(int a,int c){ 16 int fa = find(a); 17 int fc = find(c); 18 if(rrank[fa]>rrank[fc]) 19 key[fc] = fa; 20 else{ 21 key[fa] = fc; 22 if(rrank[fa]==rrank[fc]) 23 rrank[fc]++; 24 } 25 } 26 27 int num(int a){ 28 int k = find(a); 29 int ans = 0; 30 for(int i=1;i<=m;i++) 31 for(int j=1;j<=n;j++) 32 if(find(i*n+j)==k) 33 ans++; 34 35 return ans; 36 } 37 38 int main() 39 { 40 while(scanf("%d %d",&n,&m)!=EOF){ 41 if(n==0&&m==0) break; 42 for(int i=1;i<=m;i++){ 43 cin.get(); 44 for(int j=1;j<=n;j++){ 45 scanf("%c",&map[i][j]); 46 if(map[i][j]!='#') key[i*n+j] = i*n+j; 47 else key[i*n+j] = 0; 48 if(map[i][j]=='@'){//找到@的坐标 49 dx = i; 50 dy = j; 51 map[i][j] = '.'; 52 } 53 } 54 } 55 56 for(int i=1;i<m;i++){ 57 for(int j=1;j<n;j++){ 58 if(key[i*n+j]){ 59 if(key[i*n+j+1]) 60 key_union(i*n+j,i*n+j+1); 61 if(key[i*n+n+j]) 62 key_union(i*n+n+j,i*n+j); 63 } 64 } 65 if(key[i*n+n]) 66 if(key[i*n+2*n]) 67 key_union(i*n+2*n,i*n+n); 68 } 69 for(int i=1;i<n;i++) 70 if(key[m*n+i]) 71 if(key[m*n+i+1]) 72 key_union(m*n+i,m*n+i+1); 73 74 int ans = num(dx*n+dy); 75 printf("%d\n",ans); 76 } 77 }
1 //DFS解决 2 #include<iostream> 3 using namespace std; 4 5 int mov[4][2] = {-1,0,1,0,0,-1,0,1}; 6 int sum,w,h; 7 char s[21][21]; 8 9 void dfs(int x,int y){ 10 sum++;//计数 11 s[x][y] = '#'; 12 for(int i=0;i<4;++i){//四个方向前进 13 int tx = x+mov[i][0]; 14 int ty = y+mov[i][1]; 15 16 if(s[tx][ty]=='.' && tx>=0 && tx<h && ty>=0 && ty<w) 17 dfs(tx,ty);//判断该点可行后进入dfs 18 } 19 } 20 21 int main() 22 { 23 int x,y; 24 while(scanf("%d %d",&w,&h)!=EOF){ 25 if(w==0&&h==0) break; 26 for(int i=0;i<h;i++){ 27 cin.get(); 28 for(int j=0;j<w;j++){ 29 scanf("%c",&s[i][j]); 30 if(s[i][j]=='@'){//起点 31 x = i; 32 y = j; 33 } 34 } 35 } 36 sum = 0; 37 dfs(x,y); 38 printf("%d\n",sum); 39 } 40 return 0; 41 }
1 //BFS解决 2 #include<bits/stdc++.h> 3 using namespace std; 4 5 char room[23][23]; 6 int dir[4][2] = { //左上角的坐标是(0,0) 7 {-1, 0}, //向左 8 {0, -1}, //向上 9 {1, 0}, //向右 10 {0, -1} //向下 11 }; 12 13 int Wx, Hy, num; 14 #define check(x, y)(x<Wx && x>=0 && y>=0 && y<Hy) //是否在room中 15 struct node{int x, y}; 16 17 void BFS(int dx, int dy){ 18 num = 1; 19 queue<node> q; 20 node start, next; 21 start.x = dx; 22 start.y = dy; 23 q.push(start);//插入队列 24 25 while(!q.empty()){//直到队列为空 26 start = q.front();//取队首元素，即此轮循环的出发点 27 q.pop();//删除队首元素（以取出） 28 29 for(int i=0; i<4; i++){//往左上右下四个方向逐一搜索 30 next.x = start.x + dir[i][0]; 31 next.y = start.y + dir[i][1]; 32 if(check(next.x, next.y) && room[next.x][next.y]=='.'){ 33 room[next.x][next.y] = '#';//标记已经走过 34 num ++;//计数 35 q.push(next);//判断此点可行之后，插入队列，待循环判断 36 } 37 } 38 } 39 } 40 41 int main(){ 42 int x, y, dx, dy; 43 while(~scanf("%d %d",&Wx, &Hy)){ 44 if(Wx==0 && Hy==0) 45 break; 46 for(y=0; y<Hy; y++){ 47 for(x=0; x<Wx; x++){ 48 scanf("%d",&room[x][y]); 49 if(room[x][y] == '@'){//找到起点坐标 50 dx = x; 51 dy = y; 52 } 53 } 54 } 55 num = 0;//初始化 56 BFS(dx, dy); 57 printf("%d\n",num); 58 } 59 return 0; 60 }