hdu1312简单bfs

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Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15951 Accepted Submission(s): 9849

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

没啥好说的, 就bfs基本框架套进去.

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<queue>using namespace std;int w,h,cnt;char mp[21][21];struct pos{    int x,y;}st;bool vis[21][21];bool ok(int x,int y){    if(x<0 || x>=h)return false;    if(y<0 || y>=w)return false;    return true;}void bfs(){    int mve[][2]={1,0, -1,0, 0,1, 0,-1};    queue<pos> qu;    qu.push(st);    pos cur,next;    memset(vis, false, sizeof(vis));    vis[st.x][st.y] = true;    while(!qu.empty()){        cur = qu.front();qu.pop();        for(int i=0;i<4;i++){            int nx = cur.x+mve[i][0];            int ny = cur.y+mve[i][1];            if(!ok(nx,ny) || vis[nx][ny] || mp[nx][ny]=='#')continue;            vis[nx][ny]=true;            cnt++;            next.x=nx,next.y=ny;            qu.push(next);        }    }}int main(){    //freopen("hdu1312.in", "r", stdin);    //freopen("hdu.out", "w", stdout);    while(cin>>w>>h && w){        int flag=1;        cnt=1;        for(int i=0;i<h;i++){            scanf("%s", mp[i]);            if(1==flag)            for(int j=0;j<w;j++)                if(mp[i][j]=='@')                    st.x=i,st.y=j;        }        bfs();        cout<<cnt<<endl;    }    return 0;}

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