HDU 1312 Red and Black

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4306    Accepted Submission(s): 2798

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

   
   
   
   

    
    
    
    45 59 6 13
   
   
   
   

 

思路：简单的搜索题，从‘@’出发，只能站在‘.’上，不能站在‘#’上，求能踩到的.的个数（包括起始点@）

我用BFS做

#include<iostream> #include<cstdio> #include<cstring> #include<queue> using namespace std; int map[22][22]; char a[22][22]; int n,m,fx,fy,kind; queue<int> q; void Init() { int i; for(i=0;i<=n;i++) {a[i][0]='#';a[i][m+1]='#';} for(i=0;i<=m;i++) {a[0][i]='#';a[n+1][i]='#';} memset(map,0,sizeof(map)); } void bfs(int x,int y) { map[x][y]=++kind; q.push(x),q.push(y); while(!q.empty()) { x=q.front();q.pop(); y=q.front();q.pop(); if(a[x-1][y]=='.'&&!map[x-1][y]) { map[x-1][y]=++kind; q.push(x-1),q.push(y); } if(a[x+1][y]=='.'&&!map[x+1][y]) { map[x+1][y]=++kind; q.push(x+1),q.push(y); } if(a[x][y-1]=='.'&&!map[x][y-1]) { map[x][y-1]=++kind; q.push(x),q.push(y-1); } if(a[x][y+1]=='.'&&!map[x][y+1]) { map[x][y+1]=++kind; q.push(x),q.push(y+1); } } } int main() { int i,j; while(scanf("%d%d",&m,&n)!=EOF&&(n+m)) { Init(); for(i=1;i<=n;i++) { getchar(); for(j=1;j<=m;j++) { scanf("%c",&a[i][j]); if(a[i][j]=='@') { fx=i;fy=j; } } } kind=0; bfs(fx,fy); printf("%d\n",kind); } return 0; }