Warm up 2
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1881 Accepted Submission(s): 853
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.
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
题意：
题意：有水平N张牌，竖直M张牌，同一方向的牌不会相交。水平的和垂直的可能会相交，求最多能剩余多少张牌不相交，那么也就是求最少踢出去几张牌使剩下的牌都不相交。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<cstdlib>
#include<algorithm>
#include<queue>
#define INF 100000000
using namespace std;
int n,m,mp[1005][1005],mark[1005],link[1005];
struct node
{
int x,y;
}h[1005],v[1005];
bool check(int i,int j)
{
if(h[i].x<=v[j].x&&v[j].x<=h[i].x+1)
{
if(v[j].y<=h[i].y&&h[i].y<=v[j].y+1)
return true;
}
return false;
}
bool dfs(int x)
{
for(int i=1;i<=m;i++)
{
if(mark[i]==-1&&mp[x][i])
{
mark[i]=1;
if(link[i]==-1||dfs(link[i]))
{
link[i]=x;
return true;
}
}
}
return false;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==0&&m==0)
break;
memset(link,-1,sizeof(link));
memset(mp,0,sizeof(mp));
for(int i=1;i<=n;i++)
{
scanf("%d%d",&h[i].x,&h[i].y);
}
for(int i=1;i<=m;i++)
{
scanf("%d%d",&v[i].x,&v[i].y);
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if(check(i,j))
{
mp[i][j]=1;
}
}
}
int ans=0;
for(int i=1;i<=n;i++)
{
memset(mark,-1,sizeof(mark));
if(dfs(i))
ans++;
}
printf("%d
",n+m-ans);
}
return 0;
}