(bfs或者dfs)Red and Black

最新推荐文章于 2020-03-03 14:43:22 发布

是Elie呀

最新推荐文章于 2020-03-03 14:43:22 发布

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分类专栏：搜索 - DFS 搜索 - BFS 文章标签： bfs dfs

本文链接：https://blog.youkuaiyun.com/qq_37275680/article/details/79902555

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搜索 - BFS

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本文介绍了一种基于深度优先搜索（DFS）和广度优先搜索（BFS）的算法来解决在一个由黑色和红色方块组成的矩形房间中，从指定起点可达黑色方块数量的问题。输入包括房间尺寸及每个位置的颜色标记，输出为可达黑色方块总数。

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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

好久没做搜索的题目啦，直接用深搜做的，当然也可以用广搜来做，下面把代码贴在下面

DFS:

#include<iostream>
#include<cstdio>
#include<string>
#include<string.h>
using namespace std;
#define maxn 25
char mp[maxn][maxn];
int dir[4][2]={-1,0,1,0,0,-1,0,1},vis[maxn][maxn];
int w,h,cnt;

void dfs(int a,int b)
{
    vis[a][b]=1;
    cnt++;
    for(int i=0;i<4;i++)
    {
        int x=a+dir[i][0];
        int y=b+dir[i][1];
        if(x>=0&&x<h&&y>=0&&y<w&&mp[x][y]=='.'&&!vis[x][y])
        {
            dfs(x,y);
        }
    }
}

int main()
{
    int a,b;
    while(cin>>w>>h){
        if(w==0&&h==0)
            break;
        for(int i=0;i<h;i++)
        {
            for(int j=0;j<w;j++)
            {
                cin>>mp[i][j];
                if(mp[i][j]=='@')
                {
                    a=i;
                    b=j;
                }
            }
        }
        memset(vis,0,sizeof(vis));
        cnt=0;
        dfs(a,b);
        cout<<cnt<<endl;

    }
    return 0;
}

BFS:

#include<iostream>
#include<cstdio>
#include<string>
#include<string.h>
#include<queue>
using namespace std;
#define maxn 25
char mp[maxn][maxn];
int dir[4][2]={-1,0,1,0,0,-1,0,1},vis[maxn][maxn];
int w,h,cnt;
typedef struct Node{
    int m,n;
}Node;

void bfs(int a,int b)
{
    queue<Node> q;
    Node tmp,p;
    vis[a][b]=1;
    p.m=a,p.n=b;
    q.push(p);
    while(!q.empty()){
        tmp=q.front();
        cnt++;
        q.pop();
        for(int i=0;i<4;i++)
        {
            int x=tmp.m+dir[i][0];
            int y=tmp.n+dir[i][1];
            if(x>=0&&x<h&&y>=0&&y<w&&mp[x][y]=='.'&&!vis[x][y])
            {
                p.m=x,p.n=y;
                q.push(p);
                vis[x][y]=1;
            }
        }
    }
}

int main()
{
    int a,b;
    while(cin>>w>>h){
        if(w==0&&h==0)
            break;
        for(int i=0;i<h;i++)
        {
            for(int j=0;j<w;j++)
            {
                cin>>mp[i][j];
                if(mp[i][j]=='@')
                {
                    a=i;
                    b=j;
                }
            }
        }
        memset(vis,0,sizeof(vis));
        cnt=0;
        bfs(a,b);
        cout<<cnt<<endl;

    }
    return 0;
}