POJ1979/HDU 1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8336    Accepted Submission(s): 5184

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

 

题解：一道经典的搜索题。

DFS：

/*题意：'#'不可走，'.'可走，'@'为起点。      求从'@'开始走，一共可以走完多少个'.','@'也算一个。思路：dfs深度遍历周围的四个方向优化：遍历前，图的四周加边。注意：先输入高度，再输入宽。也就是先输入行数再输入列数。*/#include<cstdio>#include<cstring>#include<iostream>using namespace std;char map[25][25];int w, h;int sx, sy; //起点int ans;void dfs(int x, int y) //深度遍历{    if(map[x][y] == '#') return; //不可走，马上跳出    ans++;   //可以走 结果+1    map[x][y] = '#'; // 标记已走    dfs(x-1, y); dfs(x+1, y); //遍历周围的四个方向    dfs(x, y-1); dfs(x, y+1);}int main(){    while(scanf("%d%d", &w, &h) != EOF) //先输入列数 再输入行数    {        if(w == 0) break; // 输入0即结束        for(int i = 1; i <= h; i++)        {            for(int j = 1; j <= w; j++)            {                cin>>map[i][j];                if(map[i][j] == '@') //起点                {                    sx = i;                    sy = j;                }            }        }        for(int i = 0; i <= h+1; i++) map[i][0] = map[i][w+1] = '#'; //第0列和第w+1列加边 不可走        for(int i = 0; i <= w+1; i++) map[0][i] = map[h+1][i] = '#'; //第0行和第h+1行加边 不可走        ans = 0; //初始化结果        dfs(sx, sy);        printf("%d\n", ans);    }    return 0;}

BFS:

/************************************************************************//*  poj 1979 red and black使用方法为BFS                                                                      *//************************************************************************/#include <iostream>#include <cstring>#include <queue>using namespace std;int n,m,sx,sy,step;struct Node{int x,y;};#define MAX 25char map[MAX][MAX];int vist[MAX][MAX];int dist[4][2]={1,0,-1,0,0,1,0,-1};int inMAP(int x,int y){if(x<0 || x>=n || y<0 || y>=m) return 0;   return 1;}void BFS(){    queue <Node> Q;Node start,now,next;start.x = sx;start.y = sy;step=1;//开始位置vist[sx][sy]=1;Q.push(start);while(!Q.empty()){now = Q.front();Q.pop();for(int i=0;i<4;i++){next.x =now.x +dist[i][0];next.y =now.y +dist[i][1];//满足条件的情况if(!vist[next.x][next.y] && map[next.x][next.y]=='.' && inMAP(next.x,next.y)){   vist[next.x][next.y]=1;                  step+=1; Q.push(next);}}}}int main(){int i,j;//注意这里的行和列，被坑了。while(cin>>m>>n){     //m=6列 y ，行n=9 xmemset(vist,0,sizeof(vist));if(n+m==0)break;        for(i=0; i<n;i++)  for(j=0;j<m;j++)  cin>>map[i][j];  for(i=0;i<n;i++)  for(j=0;j<m;j++)  if(map[i][j]=='@')  {  sx=i;  sy=j;  break;  }               BFS();  cout<<step<<endl;}return 0;}

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