Red and Black(杭电oj1312）-优快云博客

本文链接：https://blog.youkuaiyun.com/xiaohang3414/article/details/50966273

本文介绍了一个迷宫寻路问题的解决方案，通过一个简单的C++程序实现了从起点出发只在黑色方块上行走，计算可达黑色方块的数量。输入为迷宫的地图布局，输出为可达的黑色方块总数。

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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)

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

Sample Output

#include <iostream>
int h,l;
char a[21][21];
using namespace std;
int f(int i,int j)
{
if(i<1||i>h||j<1||j>l) return 0;
if(a[i][j]!='#')
{
a[i][j]='#';
return 1+f(i,j+1)+f(i,j-1)+f(i+1,j)+f(i-1,j);
}
return 0;
}
int main()
{
int m,n,ans;
while(cin>>l>>h)
{
if(h==0&&l==0) break;
for(int i=1;i<=h;i++)
for(int j=1;j<=l;j++)
cin>>a[i][j];
for(int i=1;i<=h;i++)
for(int j=1;j<=l;j++)
if(a[i][j]=='@')
{m=i;n=j;}
ans=f(m,n);
cout<<ans<<endl;
}
return 0;
}

今天新学的代码不会就回来看看吧

！！