POJ 1979 Red and Black【深度优先搜索】

题目链接:http://poj.org/problem?id=1979

题目大意:一个矩形的房间地板被分为w*h个小块,每一个小块不是红的就是黑的,你首先站在一个黑色小块上,你只能朝你的四个方向(上下左右)移动,且不能到达红色的小块上,问你最多能到达多少个小块。

很简单的dfs深度优先搜索

没搜索过一个格子,将该格子设置为红色,之后的搜索就不会再搜索到该格子,就不会造成重复,因为该题有很多数据,记得每次处理数据是初始化各数组及其他数据。

代码如下:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 22
using namespace std;

int w,h;
char tile[N][N];
int ans;

void bfs(int row,int column)
{
    if(row<0||column<0||row>h||column>w)
        return ;
    else
    {
        ans++;
        if(tile[row+1][column]=='.')
        {
            tile[row+1][column]='#';
            bfs(row+1,column);
        }
        if(tile[row-1][column]=='.')
        {
            tile[row-1][column]='#';
            bfs(row-1,column);
        }
        if(tile[row][column+1]=='.')
        {
            tile[row][column+1]='#';
            bfs(row,column+1);
        }
        if(tile[row][column-1]=='.')
        {
            tile[row][column-1]='#';
            bfs(row,column-1);
        }
    }
}

int main()
{
    scanf("%d%d",&w,&h);
    int row,column;
    while(w!=0&&h!=0)
    {
        ans=0;
        for(int i=0;i<h;i++)
        scanf("%s",tile[i]);
        for(int i=0;i<h;i++)
            for(int j=0;j<w;j++)
            {
                if(tile[i][j]=='@')
                {
                    row=i;
                    column=j;
                    tile[i][j]='#';
                }
            }
        bfs(row,column);
        printf("%d\n",ans);
        memset(tile,0,sizeof(tile));
        scanf("%d%d",&w,&h);
    }

    return 0;
}

POJ 1979 Red and Black【深度优先搜索】

时间: 2024-10-19 19:43:30

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