Red and BlackTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12907Accepted Submission(s): 7983
Problem 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.#……….#.#######..#.#…..#..#.#.###.#..#.#..@#.#..#.#####.#..#…….#..#########…………11 6..#..#..#….#..#..#….#..#..###..#..#..#@…#..#..#….#..#..#..7 7..#.#….#.#..###.###…@…###.###..#.#….#.#..0 0
Sample Output
4559613
#include<iostream>#include<sstream>#include<algorithm>#include<cstdio>#include<string.h>#include<cctype>#include<string>#include<cmath>#include<vector>#include<stack>#include<queue>#include<map>#include<set>using namespace std;const int INF=40;char cnt[INF][INF];int vis[INF][INF];int dir[][2]= {{1,0},{0,1},{-1,0},{0,-1}};int ans,w,h;void dfs(int x,int y){for(int i=0; i<4; i++){int tx=dir[i][0]+x;int ty=dir[i][1]+y;if(tx>=1&&tx<=w&&ty>=1&&ty<=h&&!vis[tx][ty]&&cnt[tx][ty]=='.'){ans++;vis[tx][ty]=1;dfs(tx,ty);}}return ;}int main(){while(cin>>h>>w,h+w){memset(vis,0,sizeof(vis));int si,sj;ans=1;for(int i=1; i<=w; i++)scanf("%s",cnt[i]+1);for(int i=1; i<=w; i++){for(int j=1; j<=h; j++){if(cnt[i][j]=='@'){si=i;sj=j;break;}}}dfs(si,sj);cout<<ans<<endl;}return 0 ;}
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