HDU 1198 Farm Irrigation 并查集+枚举

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解决如何用最少的水源灌溉整个农场的问题，通过分析不同类型的水管布局，使用并查集算法确定最少水源数量。

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Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4368 Accepted Submission(s): 1888

Problem Description

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.

Output

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

Sample Input

  
   2 2
DK
HF

3 3
ADC
FJK
IHE

-1 -1

Sample Output

2
3

Author

ZHENG, Lu

Source

Zhejiang University Local Contest 2005

Recommend

Ignatius.L

题意是说给你11块田地，让它们按照不同的方式拼起来，让你求最少需要多少水源能够灌溉所有农田。

因为有的水路是连接的，所以只需要一个水源就可以了。首先将每块农田的四个方向枚举出来，然后用并查集求有几个联通分支即可。

总是为了各种各样的小错误，找很长时间的bug，心不细啊。

#include<stdio.h>
int pre[2507];
char map[57][57];
int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int n,m;

struct node
{
    bool n,s,w,e;
};
node s[11]={{1,0,1,0},{1,0,0,1},{0,1,1,0},{0,1,0,1},{1,1,0,0},{0,0,1,1},{1,0,1,1},{1,1,1,0},{0,1,1,1},{1,1,0,1},{1,1,1,1}};

void init()
{
    for(int i=0;i<n*m;i++)
        pre[i]=i;
}

bool check(int x,int y)
{
    if(x<0||x>=n||y<0||y>=m)
        return true;
    return false;
}

int find(int x)
{
    while(x!=pre[x])
        x=pre[x];
    return x;
}

void unio(int a,int b)
{
    int x=find(a);
    int y=find(b);
    if(x==y)return;
    pre[x]=y;
}

void solve(int x,int y,char ch)
{
    int a,b,xx,yy;
    a=x*m+y;
    for(int i=0;i<4;i++)
    {
        xx=x+dir[i][0];
        yy=y+dir[i][1];
        b=xx*m+yy;
        if(check(xx,yy))
            continue;
        if(i==0)
        {
            if(s[ch-'A'].n&&s[map[xx][yy]-'A'].s)
                unio(a,b);
        }
        if(i==1)
        {
            if(s[ch-'A'].s&&s[map[xx][yy]-'A'].n)
                unio(a,b);
        }
        if(i==2)
        {
            if(s[ch-'A'].w&&s[map[xx][yy]-'A'].e)
                unio(a,b);
        }
        if(i==3)
        {
            if(s[ch-'A'].e&&s[map[xx][yy]-'A'].w)
                unio(a,b);
        }
    }
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==-1&&m==-1)
            break;
        init();
        for(int i=0;i<n;i++)
            scanf("%s",map[i]);
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
                solve(i,j,map[i][j]);
        int num=0;
        for(int i=0;i<n*m;i++)
            if(pre[i]==i)
                num++;
        printf("%d\n",num);
    }
    return 0;
}