Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 22409 | Accepted: 12100 |
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
一道简单的DFS题，统计可以搜索的个数。题意：一男子站在一个黑色的瓷砖上，他旁边有黑色和红色的瓷砖，但他只能走黑色的瓷砖，而且是4个方向，问他最多能走多少个黑色的瓷砖。
#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;
int num;
char map[25][25];
int visited[25][25];
int dir[4][2] = {0, -1, 1, 0, 0, 1, -1, 0};
void DFS(int x, int y, int n, int m)
{
int mx, my;
for(int i = 0; i<4; i++)
{
mx = x+dir[i][0]; my = y+dir[i][1];
if(mx>=1 && mx<=n && my>=1 && my<=m && !visited[mx][my] && map[mx][my]==‘.‘)
{
visited[mx][my] = 1;
num++;
DFS(mx, my, n, m);
}
}
}
int main()
{
int n, m, x, y;
while(scanf("%d%d", &m, &n)!=EOF && (n || m))
{
num = 1;
memset(visited, 0, sizeof(visited));
for(int i = 1; i<=n; i++)
{
for(int j = 1; j<=m; j++)
{
cin>>map[i][j];
if(map[i][j] == ‘@‘)
{
x = i;
y = j;
visited[x][y] = 1;
}
}
}
DFS(x, y, n, m);
printf("%d\n", num);
}
return 0;
}
poj 1979 && zoj 2165 Red and Black,布布扣,bubuko.com