Red and Black
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u
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)
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
题目大意：
问最多能走‘.’多少步，‘#’是墙，不能走。
解题思路：
BFS.
代码：
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int maxN=25;
int dirI[4]={1,0,-1,0},n,m,is,js;;
int dirJ[4]={0,1,0,-1},visited[maxN][maxN];
string str[maxN];
struct node{
int i,j;
node(int i0=0,int j0=0){
i=i0,j=j0;
}
};
void bfs(){
queue <node> path;
visited[is][js]=1;
path.push(node(is,js));
int cnt=1;
while(!path.empty()){
node s=path.front();
path.pop();
for(int i=0;i<4;i++){
int di=s.i+dirI[i],dj=s.j+dirJ[i];
if(di<0||dj<0||di>=n||dj>=m||str[di][dj]=='#') continue;
if(visited[di][dj]==1) continue;
visited[di][dj]=1;
path.push(node(di,dj));
cnt++;
}
}
printf("%d\n",cnt);
}
int main(){
while(scanf("%d%d",&m,&n)!=EOF&&n&&m){
memset(visited,-1,sizeof(visited));
for(int i=0;i<n;i++){
cin>>str[i];
for(int j=0;j<m;j++){
if(str[i][j]=='@'){
is=i,js=j;
}
}
}
bfs();
}
return 0;
}