poj 2226 Muddy Fields 最小点覆盖

       转换成求最小点覆盖数

       构图方式如下：横行的连续*方块作为X集合里的顶点，纵行的连续*方块作为Y集合的顶点，若两个方块相交，则用边把这两个顶点相连。然后求最小点覆盖数（求最大匹配），即为答案。

       要注意构图后的顶点数。

 

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

const int N = 55;
const int MAX = N*N;

struct Xblock
{
	int x;
	int sty, edy;
};

struct Yblock
{
	int y;
	int stx, edx;
};

Xblock xblock[MAX];
Yblock yblock[MAX];
bool maze[MAX][MAX];
char map[N][N];
bool isvisit[MAX];
int match[MAX];

int cnt_x, cnt_y;
int row, col;

void build_graph()
{
	cnt_x = 0;
	cnt_y = 0;

	for (int i = 0; i < row; i++)
		for (int j = 0; j < col; j++)
		{
			if (map[i][j] == '*')
			{
				xblock[cnt_x].x = i;
				xblock[cnt_x].sty = j;
				while (j < col && map[i][j] == '*')
				{
					xblock[cnt_x].edy = j;
					j++;
				}
				cnt_x++;
			}
		}

	for (int j = 0; j < col; j++)
		for (int i = 0; i < row; i++)
		{
			if (map[i][j] == '*')
			{
				yblock[cnt_y].y = j;
				yblock[cnt_y].stx = i;

				while (i < row && map[i][j] == '*')
				{
					yblock[cnt_y].edx = i;
					i++;
				}
				cnt_y++;
			}
		}

	memset(maze, false, sizeof(maze));
	for (int i = 0; i < cnt_x; i++)
		for (int j = 0; j < cnt_y; j++)
			if (xblock[i].x >= yblock[j].stx && xblock[i].x <= yblock[j].edx
				&& yblock[j].y >= xblock[i].sty && yblock[j].y <= xblock[i].edy)
				maze[i][j] = true;
}

bool find(int u)
{
	for (int i = 0; i < cnt_y; i++)
	{
		if (maze[u][i] && !isvisit[i])
		{
			isvisit[i] = true;
			if (match[i] == -1 || find(match[i]))
			{
				match[i] = u;
				return true;
			}
		}
	}
	return false;
}

int main()
{
	cin >> row >> col;

	for (int i = 0; i < row; i++)
		for (int j = 0; j < col; j++)
			cin >> map[i][j];

	build_graph();

	memset(match, -1, sizeof(match));
	int ans = 0;
	for (int i = 0; i < cnt_x; i++)
	{
		memset(isvisit, false, sizeof(isvisit));
		if (find(i))
			ans++;
	}

	cout << ans << endl;
	return 0;
}