Farm Irrigation

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Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 184 Accepted Submission(s): 96

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

此题在于对11个方块的处理，上，左，右，下，为1时代表可以连通。在处理相邻方块时要改变方向，即向右移动，此方块的右边和相邻方块的左边要同时为1，才可以进行合并。

#include<cstdio>
#include<cstring>
using namespace std;

const int MAXN = 50+1;
int n, m,cnt;
char g[MAXN][MAXN];
int p[MAXN * MAXN], vis[MAXN * MAXN];
//up left right down
int way[11][4] = {{1, 1, 0, 0}, {1, 0, 1, 0}, {0, 1, 0, 1}, {0, 0, 1, 1},
					{1, 0, 0, 1}, {0, 1, 1, 0}, {1, 1, 1, 0}, {1, 1, 0, 1},
					{0, 1, 1, 1}, {1, 0, 1, 1}, {1, 1, 1, 1} };
int dir[4][2] = {{-1, 0}, {0, -1}, {0, 1}, {1, 0}};

int init(){
	for(int i = 0; i < n; i++){
		for(int j = 0; j < m; j++){
			int pos = i * m + j;
			p[pos] = pos;
			cnt++;
		}
	}
}
bool judge(int x, int y){
	if(x < 0 || x >= n || y < 0 || y >= m) return false;
	return true;
}
int get_parent(int x){
	if(p[x] != x) return get_parent(p[x]);
	else return p[x];
}
int join(int x, int y){
	int xp = get_parent(x);
	int yp = get_parent(y);
	p[xp] = yp;
	return 1;
}
int find_union(){
	for(int i = 0; i < n; i++){
		for(int j = 0; j < m; j++){
			int pos = i * m + j;
			vis[pos] = 1;
			for(int k = 0; k < 4; k++){
				int nextx = i + dir[k][0];
				int nexty = j + dir[k][1];
				int nextpos = pos + dir[k][0] * m + dir[k][1];			
   				if(!judge(nextx, nexty) || vis[nextpos]) continue;
				if(way[g[i][j] - 'A'][k] && way[g[nextx][nexty] - 'A'][(k+3) % 2]){
					if(get_parent(pos) != get_parent(nextpos)){
						cnt--;
						join(pos, nextpos);
					}
				}
			} 
		}
	}
	return 1;
}
int main(){
	while(scanf("%d%d", &n, &m) && (n > 0 && m > 0)){
		for(int i = 0; i < n; i++) scanf("%s", g[i]);
		cnt = 0;
		memset(vis, 0, sizeof(vis));
		init();
		find_union();
		printf("%d\n", cnt);
	}	
	return 1;
}