HDU 1312 red and black (BFS)

Red and black Time limit:1000 ms Memory limit:32768kb 64bit Io format:% I64d & % i64usubmit status

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 ctions, respectively. W and H are not more than 20.

There are h more lines in the data set, each of which between des 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 shocould output a line which contains the number of tiles he can reach from the initial tile (including itself ).
 

Sample Input

 6 9....#......#..............................#@...#.#..#.11 9.#..........#.#######..#.#.....#..#.#.###.#..#.#[email protected]#.#..#.#####.#..#.......#..#########............11 6..#..#..#....#..#..#....#..#..###..#..#..#@...#..#..#....#..#..#..7 7..#.#....#.#..###.###[email protected]###.###..#.#....#.#..0 0 

 

Sample output

 4559613 

 

Question:

Ask how many steps can be taken at most. '#' is a wall and cannot be taken.

Solution:

BFS.

Code:

#include<iostream>#include<cstdio>#include<queue>#include<cstring>using namespace std;const int maxN=25;int dirI[4]={1,0,-1,0},n,m,is,js;;int dirJ[4]={0,1,0,-1},visited[maxN][maxN];string str[maxN];struct node{    int i,j;    node(int i0=0,int j0=0){    i=i0,j=j0;    }};void bfs(){    queue <node> path;    visited[is][js]=1;    path.push(node(is,js));    int cnt=1;    while(!path.empty()){        node s=path.front();        path.pop();        for(int i=0;i<4;i++){            int di=s.i+dirI[i],dj=s.j+dirJ[i];            if(di<0||dj<0||di>=n||dj>=m||str[di][dj]=='#') continue;            if(visited[di][dj]==1) continue;            visited[di][dj]=1;            path.push(node(di,dj));            cnt++;        }    }    printf("%d\n",cnt);}int main(){    while(scanf("%d%d",&m,&n)!=EOF&&n&&m){        memset(visited,-1,sizeof(visited));        for(int i=0;i<n;i++){            cin>>str[i];            for(int j=0;j<m;j++){                if(str[i][j]=='@'){                    is=i,js=j;                }            }        }        bfs();    }    return 0;}

HDU 1312 red and black (BFS)