Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 26181 | Accepted: 14213 |
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
题意为：求@可以到达.（点）的最多个数， 别忘了@本身也算一个哦~！#代表不可以穿越！
大体就是这么个意思，不多说，直接上代码.......
#include <iostream>
#include <stdio.h>
#include <queue>
#include <string.h>
#define N 30
using namespace std;
int ans = 0, m, n;//ans用于计数的！
int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};//一点小技巧
int vis[N][N], map[N][N];//vis数组是用于标记是否访问过~
void bfs(int x, int y)
{
int i,ty,tx,ttx,tty;
queue<pair<int, int> > Q;//不会用queue的可以去百度看看 很好用滴
Q.push(make_pair(x, y));
vis[x][y] = 1;
while(!Q.empty())
{
tx = Q.front().first, ty = Q.front().second;
Q.pop();//别忘了一直要删除队头
for(i = 0; i < 4; i++)
{
ttx = tx + dx[i], tty = ty + dy[i];
if(ttx >= 0 && ttx < m && tty >= 0 && tty < n)
if(!vis[ttx][tty] && !map[ttx][tty])
{
vis[ttx][tty] = 1;
ans++;
Q.push(make_pair(ttx, tty));
}
}
}
}
int main()
{
int i, j, x, y;
while(~scanf("%d%d", &n, &m))
{
ans = 1;
memset(vis, 0, sizeof(vis));//比for循环清零好使
getchar();//有些人输入老是出问题，看看是不是这的原因
if(n == 0 && m == 0)
break;
char s[N];
for(i = 0; i < m; i++)
{
gets(s);
for(j = 0; j < n; j++)
{
if(s[j] == ‘.‘)
map[i][j] = 0;
else if(s[j] == ‘#‘)
map[i][j] = 1;
else
x = i, y = j;//@的坐标
}
}
bfs(x, y);
printf("%d\n", ans);
}
return 0;
}