DFS/BFS-A - Red and Black

Posted 2020-11-25 0424lrn

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

 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
",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
",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
",num);
58     }
59     return 0;
60 }