# POJ 1979 Red and Black

## Description

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.

## Input

The 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)
The end of the input is indicated by a line consisting of two zeros.

## Output

For 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

## 思路一：（DFS）  1 #include<iostream>
2 #include<cstring>
3 using namespace std;
4 const int N = 30;
5 int row, col, sx, sy;
6 int vis[N][N];
7 char a[N][N];
8 int dir = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
9 void dfs(int x, int y){
10     for(int i = 0; i < 4; i++){
11         int tx = dir[i] + x, ty = dir[i] + y;
12         if(tx < 0 || tx > row - 1 || ty < 0 || ty > col - 1)    continue;
13         if(!vis[tx][ty] && a[tx][ty] == '.'){
14             vis[tx][ty] = 1;
15             dfs(tx, ty);
16         }
17     }
18 }
19 int main(){
20     while(cin >> col >> row){
21         if(col == 0 && row == 0)    break;
22         int cnt = 0;
23         memset(vis, 0, sizeof(vis));
24         for(int i = 0; i < row; i++){
25             for(int j = 0; j < col; j++){
26                 cin >> a[i][j];
27                 if(a[i][j] == '@'){
28                     sx = i;    sy = j;
29                 }
30             }
31         }
32         vis[sx][sy] = 1;
33         dfs(sx, sy);
34         for(int i = 0; i < row; i++){
35             for(int j = 0; j < col; j++){
36                 if(vis[i][j] == 1)    cnt++;
37             }
38         }
39         cout << cnt << endl;
40     }
41     return 0;
42 }
Red and Black

## 思路二：（BFS）  1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 using namespace std;
5 struct node{
6     int x, y;
7 }que;
8 char a;
9 int book, cnt, h, w;
10 int pos = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
11 void bfs(int rx, int ry){
12     int head = 1, tail = 1;
13         que[tail].x = rx;    que[tail].y = ry;
14         book[rx][ry] = 1;
15         tail++;    cnt++;
17             for(int i = 0; i < 4; i++){
18                 int tx = que[head].x + pos[i];
19                 int ty = que[head].y + pos[i];
20                 if(tx < 0 || tx > h - 1 || ty < 0 || ty > w - 1)    continue;
21                 if(a[tx][ty] == '.' && book[tx][ty] == 0){
22                     cnt++;
23                     book[tx][ty] = 1;
24                     que[tail].x = tx;    que[tail].y = ty;
25                     tail++;
26                 }
27             }
29         }
30 }
31 int main(){
32     while(cin >> w >> h){
33         if(w == 0 && h == 0)    break;
34         memset(a, 0, sizeof(a));
35         memset(book, 0, sizeof(book));
36         cnt = 0;
37         for(int i = 0; i < h; i++){
38             cin >> a[i];
39         }
40         int rx, ry;
41         for(int i = 0; i < h; i++){
42             for(int j = 0; j < w; j++){
43                 if(a[i][j] == '@'){
44                     rx = i;
45                     ry = j;
46                 }
47             }
48         }
49         bfs(rx, ry);
50         cout << cnt << endl;
51     }
52     return 0;
53 }
Red and Black

posted @ 2019-07-16 19:58  zoom1109  阅读(...)  评论(...编辑  收藏