hdu 1312 Red and Black

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Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11962    Accepted Submission(s): 7451

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

 

//bfs#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>#include <queue>#define maxn 25;using namespace std;typedef struct{    int x, y;}node;int dx[4]={1,-1,0,0};int dy[4]={0,0,1,-1};int vis[25][25];char map[25][25];queue <node> q;int main(){    int m, n;    while (scanf("%d%d", &n, &m)!=EOF && n && m){        memset(vis, 0, sizeof(vis));        while(!q.empty()) q.pop();        int ans = 1;        node start;        for (int i = 0; i < m; i++){            getchar();            for (int j = 0; j < n; j++){                scanf("%c", &map[i][j]);                if (map[i][j] == '@'){                    start.x = i;                    start.y = j;                }            }        }        vis[start.x][start.y] = 1;        q.push(start);        while(!q.empty()){            node tmp =q.front();            q.pop();            node news = tmp;            for (int i = 0; i < 4; i++){                news.x = tmp.x + dx[i];                news.y = tmp.y + dy[i];                if (vis[news.x][news.y] == 1) continue;                if (news.x < 0 || news.x >= m || news.y < 0 || news.y >= n) continue;                if (map[news.x][news.y] == '#') continue;                if (map[news.x][news.y] == '.'){                    vis[news.x][news.y] = 1;                    ans += 1;                    q.push(news);                }            }        }        printf("%d\n", ans);    }}//dfs#include <stdio.h>#include <string.h>#include <algorithm>#include <queue>#include <math.h>#define maxn 25using namespace std;typedef struct{    int x, y;}node;char map[maxn][maxn];int vis[maxn][maxn];int dx[4]={1,-1,0,0};int dy[4]={0,0,1,-1};int m, n, ans;void dfs(int a,int b){    for (int i = 0; i < 4; i++){        int newa = a + dx[i];        int newb = b + dy[i];        if (vis[newa][newb] == 1) continue;        if (newa < 0 || newa >= m || newb < 0 || newb >= n) continue;        if (map[newa][newb] == '#') continue;        if (map[newa][newb] == '.'){            vis[newa][newb] = 1;            ans += 1;            dfs(newa, newb);            //vis[newa][newb] = 0;        }    }}int main(){    while (scanf("%d%d", &n, &m)!=EOF && n && m){        memset(vis, 0, sizeof(vis));        node start;        ans = 1;        for (int i = 0; i < m; i++){            getchar();            for (int j = 0; j < n; j++){                scanf("%c", &map[i][j]);                if (map[i][j] == '@'){                    start.x = i;                    start.y = j;                }            }        }        dfs(start.x, start.y);        printf("%d\n", ans);    }}