HDU1312 Red and Black

                                                      Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6103    Accepted Submission(s): 3887

Problem DescriptionThere 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.
 

InputThe 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)  

OutputFor 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

 

SourceAsia 2004, Ehime (Japan), Japan Domestic
 

RecommendEddy

解題思路：簡單搜尋題，深度優先搜尋，廣度優先搜尋都可以。只要做到兩點，題目即可AC。1，把所有能到達的地方走一遍，2，同一點計數不能有重複，走過後標記即可。

#include<cstdio>#include<queue>#include<string.h>using namespace std;char map[20][20];int flag[20][20];int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};int n,m;int sx,sy,maxstep=1;struct Node{    int x;    int y;};void bfs(){    queue<Node> Q;    int x,y;    int k;    Node first,next;    first.x=sx;    first.y=sy;    flag[sx][sy]=1;    Q.push(first);    while(!Q.empty())    {        first=Q.front();        Q.pop();        for(k=0;k<4;k++)        {            x=first.x+dir[k][0];            y=first.y+dir[k][1];            if(x<0||x>=n||y<0||y>=m)                continue;            if(map[x][y]=='.'&&flag[x][y]==0)            {                next.x=x;                next.y=y;                maxstep++;                Q.push(next);                flag[x][y]=1;    //記得標記啊            }        }    }}int main(){    int i,j;    while(scanf("%d%d",&m,&n)&&n||m)    {        memset(flag,0,sizeof(flag));        maxstep=1;        for(i=0;i<n;i++)        {            scanf("%s",map[i]);            for(j=0;j<m;j++)                if(map[i][j]=='@')                {                    sx=i;                    sy=j;                }        }        bfs();        printf("%d\n",maxstep);    }    return 0;}