poj1979 red andblack的两种写法（dfs bfs）_black的书写方法-优快云博客

先贴一下题目

**Red and Black **
Poj1979地址

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)
The end of the input is indicated by a line consisting of two zeros.
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

这题的意思是".“代表黑色的砖块，就是可以走的地方，”#"则是红色的砖块，就是不能走的地方，@是起点，题目让统计一下最多能走多少个砖块
这题上来一想就觉得要用dfs 毕竟暴力出奇迹再加上队里今天dhz（手搓线段树的大佬）讲了一下，我就把答案整理了一下一种是dfs 一种是bfs 由于现在实在是太晚了（一点二十二）明天还有高数课我就先贴一下代码把23333~~贴完就溜了~~ 全都是手搓的 ==

#include<bits/stdc++.h>// dfs的版本
using namespace std;
int ans;
int a[30][30];
bool vis[30][30];
int dx[]={1,-1,0,0};
int dy[]={0,0,-1,1};
int n,m;
int sx,sy;
void dfs(int x,int y) {
  for(int i=0;i<4;i++) {
    int xx=x+dx[i];
    int yy=y+dy[i];
    if(!vis[xx][yy]&&xx>=0&&xx<m&&yy>=0&&yy<n&&!a[xx][yy]) {
      vis[xx][yy]=1;
      ans++;
      dfs(xx,yy);
    }
  }
}
int main() {
  while(cin>>n>>m) {
    if(m==0&&n==0)
    break;
    ans=0;
  for(int i=0;i<m;i++)
    for(int j=0;j<n;j++) {
      char c;
      cin>>c;
      if(c=='.')
      a[i][j]=0;
      else if(c=='#')
      a[i][j]=1;
      else if(c=='@') {
        a[i][j]=1;
        sx=i;
        sy=j;
      }
    }
    dfs(sx,sy);
    cout<<ans+1<<endl;
    memset(a,0,sizeof(a));
    memset(vis,0,sizeof(vis));
  }
}

下面是bfs的版本 ~~突然感觉bfs 也有点感觉了呢！！但是没办法我还是一条咸鱼，只能等待咸鱼飞天哦不，是翻身的那一天来到hhhh~~

#include<iostream>
#include<queue>
using namespace std;
int m,n;
int sx;
int sy;
int cnt=0;
struct node {
  int x;
  int y;
};
bool a[30][30];
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
void bfs(int x,int y) {
  queue<node>q;
  node b;
  b.x=x;
  b.y=y;
  q.push(b);
  while(!q.empty()) {
    node c;
    c=q.front();
    q.pop();
    for(int i=0;i<4;i++) {
      node d;
      d.x=c.x+dx[i];
      d.y=c.y+dy[i];
      if(d.x>=0&&d.y>=0&&d.x<m&&d.y<n&&!a[d.x][d.y]) {
         q.push(d);
         a[d.x][d.y]=1;
         cnt++;
      }
    }
  }
}
int main() {
  while(cin>>n>>m) {
    if(m==0&&n==0)
    break;
    cnt=0;
  for(int i=0;i<m;i++)
    for(int j=0;j<n;j++) {
      char c;
      cin>>c;
      if(c=='.')
      a[i][j]=0;
      else if(c=='#')
      a[i][j]=1;
      else if(c=='@') {
        a[i][j]=1;
        sx=i;
        sy=j;
      }
    }
    bfs(sx,sy);
    cout<<cnt+1<<endl;
  }
}