首页 > 代码库 > Red and Black
Red and Black
Red and Black |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 118 Accepted Submission(s): 92 |
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 |
Source Asia 2004, Ehime (Japan), Japan Domestic |
Recommend Eddy |
#include<bits/stdc++.h>#define N 25using namespace std;int vis[N][N];char mapn[N][N];int dir[4][2]={ {1,0}, {-1,0}, {0,1}, {0,-1},};int cur=0;int n,m;bool ok(int x,int y){ if(x<0||x>=m||y<0||y>=n||vis[x][y]||mapn[x][y]==‘#‘) return false; return true;}void dfs(int x,int y){ cur++; for(int i=0;i<4;i++) { int fx=x+dir[i][0]; int fy=y+dir[i][1]; if(ok(fx,fy)) { vis[fx][fy]=1; dfs(fx,fy); } }}int main(){ //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin); while(scanf("%d%d",&n,&m)!=EOF&&n&&m) { memset(vis,0,sizeof vis); int x,y; getchar(); for(int i=0;i<m;i++) { scanf("%s",mapn[i]); //cout<<mapn[i]<<endl; for(int j=0;j<n;j++) { if(mapn[i][j]==‘@‘) { x=i; y=j; } //cout<<mapn[i][j]; } //cout<<endl; } cur=0; vis[x][y]=1; dfs(x,y); printf("%d\n",cur); } return 0;}
Red and Black
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。