hdu 1198Farm Irrigation

原创于 2018-04-03 21:59:17 发布 · 146 阅读

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本文介绍了一种使用并查集算法解决农田灌溉问题的方法。通过计算最少需要多少个水源点来确保整个农田得到充分灌溉。利用11种不同类型的水管布局，将二维地图转化为一维数组进行处理。

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.

Input

There are several test cases! In each test case, the first line contains 2 integers M and N, then M lines follow. In each of these lines, there are N characters, in the range of 'A' to 'K', denoting the type of water pipe over the corresponding square. A negative M or N denotes the end of input, else you can assume 1 <= M, N <= 50.

<b< dd="">

Output

For each test case, output in one line the least number of wellsprings needed.

<b< dd="">

Sample Input

2 2
DK
HF

3 3
ADC
FJK
IHE

-1 -1

Sample Output

2
3

题解

对于这道题，就是找有多少个连通分量，用并查集可以解决这道题

有一个巧妙的地方，二维数组可以转换成一位数组

#include <iostream>
#include <cstdio>
using namespace std;
int n,m,pre[550*550+1],cnt;
char map[550][550];
const bool vis=true;
int dir[11][4]={
    {1,1,0,0},{0,1,1,0},{1,0,0,1},{0,0,1,1},{0,1,0,1},{1,0,1,0},
    {1,1,1,0},{1,1,0,1},{1,0,1,1},{0,1,1,1},{1,1,1,1}
};//按左上右下四个方向，1表示有水管，0表示没有水管
void init(int n)
    {
        for (int i=1;i<=n;i++)
            pre[i]=i;
        cnt=n;
    }
int find(int x)
    {
        if(x==pre[x]) return x;
        else return pre[x]=find(pre[x]);
    }
void merge(int ax,int ay,int bx,int by,bool deti)
    {
        if(bx>n||by>m) return ;
        int at=map[ax][ay]-'A';
        int bt=map[bx][by]-'A';
        int flag=0;
        if(deti)//判断竖直方向
            {
                if(dir[at][3]&&dir[bt][1])
                    flag=1;
            }
        else//判断水平方向
            {
                if(dir[at][2]&&dir[bt][0])
                    flag=1;
            }
        if(flag)
            {
                int xx=find((ax-1)*m+ay);//将二维的数组转化成一位的数组
                int yy=find((bx-1)*m+by);
                if(xx!=yy)
                    {
                        pre[xx]=yy;
                        cnt--;
                    }
            }
    }
int main()
    {
        while (cin>>n>>m)
            {
                if(n==-1&&m==-1) break;
                for (int i=1;i<=n;i++)
                   //scanf("%s",map[i]+1);//这个就是对二维数组第二个变量的位置地址加上一个1，因为是从1开始的
                   for (int j=1;j<=m;j++)
                        cin>>map[i][j];
                init(n*m);
                for (int i=1;i<=n;i++)
                    for (int j=1;j<=m;j++)
                        {
                            merge(i,j,i,j+1,!vis);//横向进行判断
                            merge(i,j,i+1,j,vis);//纵向判断
                        }
                cout<<cnt<<endl;
            }
        return 0;
    }