hdu1198 不一样的搜索水题

最新推荐文章于 2018-02-08 12:45:08 发布

原创最新推荐文章于 2018-02-08 12:45:08 发布 · 446 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#acm #算法 #搜索

ACM 同时被 3 个专栏收录

37 篇文章

订阅专栏

-------------图论-------------

25 篇文章

订阅专栏

搜索

12 篇文章

订阅专栏

本文介绍了一个农场灌溉模拟问题，通过不同的水管布局确保所有农田得到灌溉。采用深度优先搜索算法计算最少所需的水源数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Benny has a spacious farm land to irrigate. The farm land is a rectangle, and is divided into a lot of samll squares. Water pipes are placed in these squares. Different square has a different type of pipe. There are 11 types of pipes, which is marked from A to K, as Figure 1 shows.

Figure 1

Benny has a map of his farm, which is an array of marks denoting the distribution of water pipes over the whole farm. For example, if he has a map

ADC
FJK
IHE

then the water pipes are distributed like

Figure 2

Several wellsprings are found in the center of some squares, so water can flow along the pipes from one square to another. If water flow crosses one square, the whole farm land in this square is irrigated and will have a good harvest in autumn.

Now Benny wants to know at least how many wellsprings should be found to have the whole farm land irrigated. Can you help him?

Note: In the above example, at least 3 wellsprings are needed, as those red points in Figure 2 show.

题意 A--K对应不同水管要事整个田地都能灌溉到水

问需要到少水源一开始想的就是深搜只是方向都不同罢了和求连通区域个数没什么区别属于最水的搜索题

ac code

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;
const int maxn=60;
char mat[maxn][maxn];
bool vis[maxn][maxn];
int row,col;
int dir[12][4]=
{
    {1,0,0,1},{1,1,0,0},{0,0,1,1},{0,1,1,0},
    {1,0,1,0},{0,1,0,1},{1,1,0,1},{1,0,1,1},
    {0,1,1,1},{1,1,1,0},{1,1,1,1}
};
int dx[]= {-1,0,1,0};
int dy[]= {0,1,0,-1};

bool judge(int x,int y)
{
    return x>0&&x<=row&&y>0&&y<=col&&!vis[x][y];
}

void dfs(int from,int x,int y)
{
    int num=mat[x][y]-'A',xx,yy;
    if(from<4)
        if(!dir[num][(from+2)%4]){vis[x][y]=false; return ;}
    for(int i=0; i<4; i++)
    {
        if(dir[num][i])
        {
            xx=x+dx[i];
            yy=y+dy[i];
            if(judge(xx,yy))
            {
                vis[xx][yy]=true;
                dfs(i,xx,yy);
            }
        }
    }
}

int main()
{
    while(scanf("%d%d",&row,&col),row>0&&col>0)
    {
        getchar();
        for(int i=1; i<=row; i++)
            scanf("%s",mat[i]+1);
        memset(vis,false,sizeof vis);
        int ans=0;
        for(int i=1; i<=row; i++)
            for(int j=1; j<=col; j++)
                if(!vis[i][j])
                {
                    ans++;
                    vis[i][j]=true;
                    dfs(maxn,i,j);
                }
        printf("%d\n",ans);
    }
    return 0;
}