POJ 1979 Red and Black

Red and Black

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 22915 Accepted: 12361

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)
The end of the input is indicated by a line consisting of two zeros.

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

第一道广搜题目；

题目很简单就是求@能到达的 ‘.’的最大个数；

运用队列和结构体每次搜索队首。然后将队首清除。循环找不到停止。

输出值即可；

代码如下：

#include<cstdio>#include<iostream>#include<queue>#include<algorithm>using namespace std;int n,m,ma;char tian[1005][1005];struct node{    int x,y;};bool cp(node q){    if(q.x<0||q.x>=n||q.y<0||q.y>=m||tian[q.x][q.y]=='#')       return false;    return true;}queue<node> q;int dir[4][2]={1,0,0,1,-1,0,0,-1};void bfs(int x,int y){    tian[x][y]='#';    node t,te;    int i;    t.x=x,t.y=y;    while(!q.empty())        q.pop();        q.push(t);    while(!q.empty())    {        t=q.front();        q.pop();        for(i=0;i<4;i++)        {            te.x=t.x+dir[i][0];            te.y=t.y+dir[i][1];            if(cp(te))            {                ma++;                tian[te.x][te.y]='#';                q.push(te);            }        }    }}int main(){    int i,j;    while(~scanf("%d%d",&m,&n))    {        if(m==0&&n==0)            break;        for(i=0;i<n;i++)        {            scanf("%s",&tian[i]);        }        for(i=0;i<n;i++)            for(j=0;j<m;j++)        {            if(tian[i][j]=='@')            {                ma=1;                bfs(i,j);            }        }        printf("%d\n",ma);    }    return 0;}