HDU 4619 Warm up 2 最大匹配数

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本文介绍了一种解决水平和垂直骨牌放置问题的算法，通过二分匹配找到最少移除骨牌的数量，使得剩余骨牌不重叠。提供了一个完整的C++实现案例。

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Warm up 2

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 967 Accepted Submission(s): 465

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

Sample Output

4
6

Source

2013 Multi-University Training Contest 2

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zhuyuanchen520

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，求最少踢出去几张牌使剩下的牌都不相交。

二分匹配最小点覆盖=最大匹配。

#include<stdio.h>
#include<string.h>
using namespace std;
int n,m;
int g[1007][1007],link[1007];
bool vis[1007];
struct node
{
    int x,y;
}h[1007],v[1007];
bool judge(int i,int j)
{
    //注释掉的也能判断是否重合
    /*if(h[i].x==v[j].x||h[i].x+1==v[j].x)
    {
        if(v[j].y==h[i].y||v[j].y+1==h[i].y)
        return true;
    }*/
    //判断是否重合
    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 find(int i)
{
    for(int j=1;j<=m;j++)
    if(g[i][j]&&!vis[j])
    {
        vis[j]=true;
        if(link[j]==0||find(link[j]))
        {
            link[j]=i;
            return true;
        }
    }
    return false;
}
int main()
{
    while(scanf("%d%d",&n,&m),n|m)
    {
        int count=0;
        memset(g,0,sizeof(g));
        memset(link,0,sizeof(link));
        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(judge(i,j))
        {
            g[i][j]=1;
        }
        for(int i=1;i<=n;i++)
        {
            memset(vis,false,sizeof(vis));
            if(find(i))
            count++;
        }
        printf("%d\n",n+m-count);
    }
    return 0;
}