HDOJ 1312 Red and Black（BFS）

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
////  main.cpp//  red and black////  Created by 张嘉韬 on 16/3/25.//  Copyright © 2016年 张嘉韬. All rights reserved.//#include <iostream>#include <cstring>using namespace std;int dx[]={-1,1,0,0};int dy[]={0,0,-1,1};int m,n,map[30][30],book[30][30],sx,sy,counter;struct node{    int x;    int y;}nodes[500];int safe(int x,int y){    int flag=1;    if(x<=0||x>n||y<=0||y>m||map[x][y]==0||book[x][y]==1) flag=0;    return flag;}void bfs(int x,int y){    int head,tail;    head=tail=1;    nodes[tail].x=x;    nodes[tail].y=y;    book[x][y]=1;    tail++;    while(head<tail)    {        int tempx,tempy;        for(int i=0;i<4;i++)        {            tempx=nodes[head].x+dx[i];            tempy=nodes[head].y+dy[i];            if(safe(tempx,tempy)==1)            {                nodes[tail].x=tempx;                nodes[tail].y=tempy;                book[tempx][tempy]=1;                tail++;                counter++;            }        }        head++;    }    }int main(int argc, const char * argv[]) {    //freopen("/Users/zhangjiatao/Desktop/input.txt","r",stdin);    while(scanf("%d%d",&m,&n)==2)    {        if(m==0&&n==0) return 0;        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                char temp;                cin>>temp;                if(temp=='.') map[i][j]=1;                else if(temp=='#') map[i][j]=0;                else sx=i,sy=j;                book[i][j]=0; //0 is avil ,1 is used            }        }        counter=0;        bfs(sx,sy);        cout<<counter+1<<endl;    }    return 0;}

总结
这次写的其实挺顺利的，主要是按照了书上有良好的代码风格，以后也要注意，自己思考写下固然很重要但不可否认的是自己写的一套总是出现一些细节上的小问题，这些小问题反而不好找出来，所以，在自己思考吃透了算法的思想可以照着优秀的代码去构建自己的代码风格