Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 41057 | Accepted: 22269 |
Description
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Input
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
Sample Input
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
題目的大概意思就是一個勇士從某一點(@)出發,只能走相鄰的黑色的格子(#),而不能走紅色格子(.)
代碼:
//
// main.cpp
// DFS——紅與黑.1
//
// Created by showlo on 2018/4/19.
// Copyright © 2018年 showlo. All rights reserved.
//
#include <stdio.h>
#include <algorithm>
using namespace std;
#define max 25
char map[max][max];
int M,N,sx,sy;
int num;
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
//void dfs()
void dfs(int x,int y){
int i,nx,ny;
if (x<0||x>=M||y<0||y>=N) {
return;
}
for (i=0; i<4; i++) {
nx=x+dx[i];
ny=y+dy[i];
//printf("%d %d\n",nx,ny);
if(map[nx][ny]=='.') {
num++;
map[nx][ny]='#';
dfs(nx, ny);
}
}
return;
}
int main() {
int i,j;
while (scanf("%d %d",&N,&M)!=EOF) {
if (M==0||N==0) {
break;
}
memset(map, 0, sizeof(map));
num=0;
for (i=0; i<M; i++) {
scanf("%s",map[i]);
}
for (i=0; i<M; i++) {
for (j=0; j<N; j++) {
if (map[i][j]=='@') {
sx=i;
sy=j;
}
}
}
// printf("%d %d\n",sx,sy);
num=1;
map[sx][sy]='#';
dfs(sx,sy);
printf("%d\n",num);
}
return 0;
}