hdu4619Warm up 2

Problem Description

　　Some 1×2 dominoes are placed on a plane. Each dominoe is placed either horizontally or vertically. It's guaranteed the dominoes in the same direction are not overlapped, but horizontal and vertical dominoes may overlap with each other. You task is to remove some dominoes, so that the remaining dominoes do not overlap with each other. Now, tell me the maximum number of dominoes left on the board.

 

Input

　　There are multiple input cases.
　　The first line of each case are 2 integers: n(1 <= n <= 1000), m(1 <= m <= 1000), indicating the number of horizontal and vertical dominoes.
Then n lines follow, each line contains 2 integers x (0 <= x <= 100) and y (0 <= y <= 100), indicating the position of a horizontal dominoe. The dominoe occupies the grids of (x, y) and (x + 1, y).
　　Then m lines follow, each line contains 2 integers x (0 <= x <= 100) and y (0 <= y <= 100), indicating the position of a horizontal dominoe. The dominoe occupies the grids of (x, y) and (x, y + 1).
　　Input ends with n = 0 and m = 0.

 

Output

　　For each test case, output the maximum number of remaining dominoes in a line.

 

Sample Input

   
   
   
   

    
    
    
    2 3
0 0
0 3
0 1
1 1
1 3
4 5
0 1
0 2
3 1
2 2
0 0
1 0
2 0
4 1
3 2
0 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    4
6
   
   
   
   

 

Source

2013 Multi-University Training Contest 2

 

Recommend


分别把竖着的，横着的牌分成2分，把有重叠的牌连一条边，进行二分匹配，然后求最大独立集=顶点数-最大匹配数

#include<map>
#include<set>
#include<list>
#include<stack>
#include<queue>
#include<vector>
#include<cmath>
#include<cstdio>
#include<string>
#include<iostream>
#include<iterator>
#include<algorithm>

using namespace std;

const int maxn=1005;

struct node
{
	int to;
	int next;
}edge[maxn*maxn];
int head[maxn];
int tot;
int mark[maxn];
bool used[maxn];
int n,m;

struct List
{
	int x1,y1;
	int x2,y2;
}H_point[maxn],V_point[maxn];

void addedge(int from,int to)
{
	edge[tot].to=to;
	edge[tot].next=head[from];
	head[from]=tot++;
}

bool is_overlapped(int a,int b)
{
	if(H_point[a].x1==V_point[b].x1 && H_point[a].y1==V_point[b].y1)
	  return true;
    else if(H_point[a].x1==V_point[b].x2 && H_point[a].y1==V_point[b].y2)
      return true;
    else if(H_point[a].x2==V_point[b].x1 && H_point[a].y2==V_point[b].y1)
      return true;
    else if(H_point[a].x2==V_point[b].x2 && H_point[a].y2==V_point[b].y2)
      return true;
    return false;
}

bool dfs(int x)
{
	for(int i=head[x];i!=-1;i=edge[i].next)
	{
		if(!used[edge[i].to])
		{
			used[edge[i].to]=1;
			if(mark[edge[i].to]==-1 || dfs(mark[edge[i].to]))
			{
				mark[edge[i].to]=x;
				return true;
			}
		}
	}
	return false;
}

int hungary()
{
	memset(mark,-1,sizeof(mark));
	int ans=0;
	for(int i=1;i<=n;i++)
	{
		memset(used,0,sizeof(used));
		if(dfs(i))
		  ans++;
	}
	return ans;
}

int main()
{
	while(~scanf("%d%d",&n,&m))
	{
		if(n==0 && m==0)
		  break;
		for(int i=1;i<=n;i++)
		{
			scanf("%d%d",&H_point[i].x1,&H_point[i].y1);
			H_point[i].x2=H_point[i].x1+1;
			H_point[i].y2=H_point[i].y1;
		}
		for(int i=1;i<=m;i++)
		{
			scanf("%d%d",&V_point[i].x1,&V_point[i].y1);
			V_point[i].x2=V_point[i].x1;
			V_point[i].y2=V_point[i].y1+1;
		}
		memset(head,-1,sizeof(head));
		tot=0;
		for(int i=1;i<=n;i++)
		  for(int j=1;j<=m;j++)
		  {
  			 if(is_overlapped(i,j))
 			  	addedge(i,j);
  		  }
		printf("%d\n",m+n-hungary());
	}
	return 0;
}