思路:從*所在的位置出發,dfs歷遍所有與它聯通的點並將其變爲#
dfs實現關鍵思路在於兩點:
1.哪些點是與當前點聯通的
2.dfs返回條件
代碼如下:
#include<iostream>
using namespace std;
char maze[35][90];
int visit[35][90];
int r;//the row of maze
void DFS(int row, int col) {
if (row<0 || row>r - 1)return;
int len =strlen(maze[row]);
if (col<0 || col>len - 1)return;
if (visit[row][col]||maze[row][col] == 'X')return;
visit[row][col] = 1;
if (maze[row][col] == ' '|| maze[row][col] == '*')maze[row][col] = '#';
for (int i = -1; i <= 1; i++)
for (int j = -1; j <= 1; j++)
{
if (i == 0 && j == 0)continue;
DFS(row + i, col + j);
}
}
int main() {
//FILE*stream;
//freopen_s(&stream, "C:\\Data\\784Maze.txt", "r", stdin);
int num;
scanf_s("%d", &num);
char c = getchar();
while (num--) {
//read in maze
r = 0;
while (1) {
gets_s(maze[r]);
if (maze[r][0] == '_')break;
r++;
}
//find *
int x, y;
for (int i = 0; i < r; i++)
for (int j = 0; j < strlen(maze[i]); j++)
if (maze[i][j] == '*') {
x = i; y = j; break;
}
memset(visit, 0, sizeof(visit));
DFS(x, y);
//output
for (int i = 0; i <= r; i++)
printf_s("%s\n", maze[i],90);
}
}