Hdu 1507 Uncle Tom's Inherited Land* 分类： Brush Mode 2014-07-30 09:28 112人阅读 评论(0) 收藏

Uncle Tom's Inherited Land*

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1869    Accepted Submission(s): 777
Special Judge

Problem Description

Your old uncle Tom inherited a piece of land from his great-great-uncle. Originally, the property had been in the shape of a rectangle. A long time ago, however, his great-great-uncle decided to divide the land into a grid of small squares. He turned some of the squares into ponds, for he loved to hunt ducks and wanted to attract them to his property. (You cannot be sure, for you have not been to the place, but he may have made so many ponds that the land may now consist of several disconnected islands.)

Your uncle Tom wants to sell the inherited land, but local rules now regulate property sales. Your uncle has been informed that, at his great-great-uncle's request, a law has been passed which establishes that property can only be sold in rectangular lots the size of two squares of your uncle's property. Furthermore, ponds are not salable property.

Your uncle asked your help to determine the largest number of properties he could sell (the remaining squares will become recreational parks).

 

Input

Input will include several test cases. The first line of a test case contains two integers N and M, representing, respectively, the number of rows and columns of the land (1 <= N, M <= 100). The second line will contain an integer K indicating the number of squares that have been turned into ponds ( (N x M) - K <= 50). Each of the next K lines contains two integers X and Y describing the position of a square which was turned into a pond (1 <= X <= N and 1 <= Y <= M). The end of input is indicated by N = M = 0.

 

Output

For each test case in the input your program should first output one line, containing an integer p representing the maximum number of properties which can be sold. The next p lines specify each pair of squares which can be sold simultaneity. If there are more than one solution, anyone is acceptable. there is a blank line after each test case. See sample below for clarification of the output format.

 

Sample Input

4 4
6
1 1
1 4
2 2
4 1
4 2
4 4
4 3
4
4 2
3 2
2 2
3 1
0 0

 

Sample Output

4
(1,2)--(1,3)
(2,1)--(3,1)
(2,3)--(3,3)
(2,4)--(3,4)

3
(1,1)--(2,1)
(1,2)--(1,3)
(2,3)--(3,3)

题意就是找出最多的1*2的方块来，UP主的主要思路为：以i+j的奇偶性为分类，因为每个1*2的方块肯定是由一个奇点和一个偶点构成，二分图是思路都是某个东西由2类东西构成，再去用匈牙利算法，然后再用map数组偶点存匹配的方向

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int map[105][105];
bool vis[105][105];
int d[5][2]={{0,0},{1,0},{0,1},{-1,0},{0,-1}};
int x, y;
void myscanf()
{
	memset(map,0,sizeof(map));
	int m, a, b;
	scanf("%d",&m);
	while(m--)
	{
		scanf("%d%d",&a,&b);
		map[a][b] = -1;
	}
}

void myprintf(int xi, int yi)
{
	int f = map[xi][yi];
	int fx = d[f][0] + xi;
	int fy = d[f][1] + yi;
	printf("(%d,%d)--(%d,%d)
",xi,yi,fx,fy);
}

bool pd(int xi, int yi)
{
	if(xi<=0 || xi>x || yi<=0 || yi>y || map[xi][yi]==-1)	return false;
	return true;
}

bool dfs(int xi,int yi)
{
	for(int i=1;i<=4;i++)
	{
		int fx = xi+d[i][0];
		int fy = yi+d[i][1];
		if(pd(fx,fy) && !vis[fx][fy])
		{
			vis[fx][fy] = true;
			int f = map[fx][fy];
			int ex = fx + d[f][0];
			int ey = fy + d[f][1];
			if(!f || dfs(ex,ey))
			{
				map[fx][fy] = (i+2)%4==0?4:(i+2)%4;
				return true;
			}
		}
	}
	return false;
}


int find()
{
	int res = 0;
	for(int i=1;i<=x;i++)
	{
		for(int j=1;j<=y;j++)
		{
			if((i+j)%2)
			{
				memset(vis,0,sizeof(vis));
				if(map[i][j]==-1)	continue;
				if(dfs(i,j))	res++;
			}
		}
	}
	return res;
}

int main()
{
//	freopen("in.txt","r+",stdin);
//	freopen("out1.txt","w+",stdout);
	while(scanf("%d%d",&x,&y)!=EOF)
	{
		if(x==0 && y==0)	break;
		myscanf();
		cout<<find()<<endl;
		for(int i=1;i<=x;i++)
		{
			for(int j=1;j<=y;j++)
			{
				if(map[i][j]>0)
				{
					myprintf(i,j);
				}
			}
		}
		printf("
");
	}
//	fclose(stdin);
//	fclose(stdout);
	return 0;
}

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