DFS/BFS-A - Red and Black

A - Red and Black

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 Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample 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

 //并查集解决
 #include<iostream>
 using namespace std;

 const int h = ;
 char map[h][h];
 int  key[h*h];
 int rrank[h*h];
 int  n,m,dx,dy;

 int find(int a){
     return a==key[a]? a : key[a]=find(key[a]);
 }

 void key_union(int a,int c){
     int fa = find(a);
     int fc = find(c);
     if(rrank[fa]>rrank[fc])
         key[fc] = fa;
     else{
         key[fa] = fc;
         if(rrank[fa]==rrank[fc])
             rrank[fc]++;
     }
 }

 int num(int a){
     int k = find(a);
     int ans = ;
     for(int i=;i<=m;i++)
         for(int j=;j<=n;j++)
             if(find(i*n+j)==k)
                 ans++;

     return ans;
 }

 int main()
 {
     while(scanf("%d %d",&n,&m)!=EOF){
         if(n==&&m==)    break;
         for(int i=;i<=m;i++){
             cin.get();
             for(int j=;j<=n;j++){
                 scanf("%c",&map[i][j]);
                 if(map[i][j]!='#')    key[i*n+j] = i*n+j;
                 else                key[i*n+j] = ;
                 if(map[i][j]=='@'){//找到@的坐标
                     dx = i;
                     dy = j;
                     map[i][j] = '.';
                 }
             }
         }

         for(int i=;i<m;i++){
             for(int j=;j<n;j++){
                 if(key[i*n+j]){
                     if(key[i*n+j+])
                         key_union(i*n+j,i*n+j+);
                     if(key[i*n+n+j])
                         key_union(i*n+n+j,i*n+j);
                 }
             }
             if(key[i*n+n])
                 if(key[i*n+*n])
                     key_union(i*n+*n,i*n+n);
         }
         for(int i=;i<n;i++)
             if(key[m*n+i])
                 if(key[m*n+i+])
                     key_union(m*n+i,m*n+i+);    

         int ans = num(dx*n+dy);
         printf("%d\n",ans);
     }
 }

 //DFS解决
 #include<iostream>
 using namespace std;

 int mov[][] = {-,,,,,-,,};
 int sum,w,h;
 char s[][];

 void dfs(int x,int y){
     sum++;//计数
     s[x][y] = '#';
     for(int i=;i<;++i){//四个方向前进
         int tx = x+mov[i][];
         int ty = y+mov[i][];

         if(s[tx][ty]=='.' && tx>= && tx<h && ty>= && ty<w)
             dfs(tx,ty);//判断该点可行后进入dfs
     }
 }

 int main()
 {
     int x,y;
     while(scanf("%d %d",&w,&h)!=EOF){
         if(w==&&h==)    break;
         for(int i=;i<h;i++){
             cin.get();
             for(int j=;j<w;j++){
                 scanf("%c",&s[i][j]);
                 if(s[i][j]=='@'){//起点
                     x = i;
                     y = j;
                 }
             }
         }
         sum = ;
         dfs(x,y);
         printf("%d\n",sum);
     }
     return ;
 }

 //BFS解决
 #include<bits/stdc++.h>
 using namespace std;

 char room[][];
 int dir[][] = { //左上角的坐标是(0,0)
     {-, },     //向左
     {, -},     //向上
     {, },     //向右
     {, -}        //向下
 };

 int Wx, Hy, num;
 #define check(x, y)(x<Wx && x>=0 && y>=0 && y<Hy)    //是否在room中
 struct node{int x, y};

 void BFS(int dx, int dy){
     num = ;
     queue<node> q;
     node start, next;
     start.x = dx;
     start.y = dy;
     q.push(start);//插入队列 

     while(!q.empty()){//直到队列为空
         start = q.front();//取队首元素，即此轮循环的出发点
         q.pop();//删除队首元素（以取出） 

         for(int i=; i<; i++){//往左上右下四个方向逐一搜索
             next.x = start.x + dir[i][];
             next.y = start.y + dir[i][];
             if(check(next.x, next.y) && room[next.x][next.y]=='.'){
                 room[next.x][next.y] = '#';//标记已经走过
                 num ++;//计数
                 q.push(next);//判断此点可行之后，插入队列，待循环判断
             }
         }
     }
 } 

 int main(){
     int x, y, dx, dy;
     while(~scanf("%d %d",&Wx, &Hy)){
         if(Wx== && Hy==)
             break;
         for(y=; y<Hy; y++){
             for(x=; x<Wx; x++){
                 scanf("%d",&room[x][y]);
                 if(room[x][y] == '@'){//找到起点坐标
                     dx = x;
                     dy = y;
                 }
             }
         }
         num = ;//初始化
         BFS(dx, dy);
         printf("%d\n",num);
     }
     return ;
 }