杭电1198 Farm Irrigation（并查集）

最新推荐文章于 2021-10-25 19:11:05 发布

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最新推荐文章于 2021-10-25 19:11:05 发布

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本文探讨了如何通过最小化井喷数量来实现农田灌溉系统的全面覆盖，结合给定的地图信息，采用图论的方法进行分析，旨在找到最有效的灌溉策略，确保农田在秋季获得丰收。

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Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5775 Accepted Submission(s): 2505

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/*
题目大意：有些地块有的方向水管可以和外边联通，求最少需要的管子数量
终于搞定了，也没那么难啊，自己把自己吓住了，只要把X或Y一个方向的当成自己的父节点就可以，合并和查询都找那个方向的
Time:2014-8-9 22:15
*/
#include<stdio.h>
#include<string.h>
struct Node
{
	bool u;
	bool d;
	bool l;
	bool r;
}p[]=
{
 {1, 0, 1, 0,}, //A  
 {1, 0, 0, 1,},//B   
 {0, 1, 1, 0,},//C  
 {0, 1, 0, 1,},//D  
 {1, 1, 0, 0,},//E  
 {0, 0, 1, 1,},//F  
 {1, 0, 1, 1,},//G  
 {1, 1, 1, 0,},//H  
 {0, 1, 1, 1,},//I  
 {1, 1, 0, 1,},//J  
 {1, 1, 1, 1}//K
};//11种类型 
 int X,Y;
int  father[600*600];
int map[600][600];
int findRoot(int a)
{
	return father[a]==a?a:father[a]=findRoot(father[a]);
}
void merge(int a,int b)
{
	a=findRoot(a);
	b=findRoot(b);
	if(a!=b)
	{
		father[a]=b;
	}
}
void Init()
{
		for(int i=0;i<=X;i++)
		{
			for(int j=0;j<=Y;j++)
			{
				father[i*Y+j]=i*Y+j;
				map[i][j]=0;
			}
		}
}
int main()
{
	while(scanf("%d%d",&X,&Y)!=EOF)
	{
		if(X==-1||Y==-1)
			break;
		char ch[600];
		Init();
		for(int i=0;i<X;i++)
		{	
			scanf("%s",ch);
			for(int j=0;j<Y;j++)
			{
			map[i][j]=ch[j]-'A';
			}
		}
		/*
		for(int i=0;i<X;i++){
		for(int j=0;j<Y;j++)
		printf("%d",map[i][j]);
		printf("\n");
		}
		*/
		for(int i=0;i<X;i++){
			for(int j=0;j<Y;j++){
			//	printf("fa=%d %d",father[i*Y+j],i*Y+j);
				if(p[map[i][j]].r&&p[map[i][j+1]].l&&j+1<Y){
				merge(i*Y+j,i*Y+j+1);
				//printf("i=%d j=%d i=%d j+1=%d\n",i,j,i,j+1);
				}
				if(p[map[i][j]].d&&p[map[i+1][j]].u&&i+1<X){
				merge(i*Y+j,(i+1)*Y+j);
				//printf("i=%d j=%d i=%d j+1=%d\n",i+1,j,i,j);
				}
			}
		}
		int num=0;
		for(int i=0;i<X;i++)
		for(int j=0;j<Y;j++){
			if(father[i*Y+j]==i*Y+j)
			num++;
			//printf("f %d=%d\n",father[i*Y+j],i*Y+j);
		}
		printf("%d\n",num);
	}
return 0;
}