hdu 4619—二分图求最大独立集

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

 

这道题的题意是给你一些1*2的骨牌，骨牌可能是垂直放置所占坐标是（x，y).(x,y+1) , 也可能是平行放置所占坐标是（x，y）。（x+1，y），骨牌不会重合，但是骨牌的坐标可能重合，叫你将这些坐标重合的骨牌去掉，问留下骨牌数最多为多少？？？

这道题就是一道最大独立集的题目，这里我们要建图，以横起的和竖起的骨牌建图，就是吧坐标重合的那些点相连，最后总点数减去最大二分匹配数

#include<stdio.h>

#include<iostream>

#include<algorithm>

#include<string.h>

#include<math.h>

using namespace std;

#define ll long long

#define nn 2005

int ma[nn][nn];

int use[nn];

int id[nn];

int n,m;

int dfs(int x)

{

    int i;

    for(i=1;i<=m+n;i++)

    {

        if(use[i]==false&&ma[x][i]==1)

        {

            use[i]=true;

            if(id[i]==-1||dfs(id[i]))

            {

                id[i]=x;

                return true;

            }

        }

    }

  return false;

}

int cnt;

void fin()

{

    memset(id,-1,sizeof(id));

    int i;

    cnt=0;

    for(i=1;i<=n+m;i++)

    {

        memset(use,false,sizeof(use));

        if(dfs(i))

            cnt++;

    }

    printf("%d\n",(n+m)-cnt);

}

 struct node

 {

     int x,y;

     int num;

 }point[nn];

 int main()

 {

     int i,j;

     while(scanf("%d %d",&n,&m)&&m+n)

     {

         memset(ma,0,sizeof(ma));

         for(i=1;i<=n+m;i++)

         {

             scanf("%d %d",&point[i].x,&point[i].y);

             point[i].num=i;

         }

         for(i=1;i<=n;i++)

         {

             for(j=n+1;j<=m+n;j++)

             {

                    if( ((point[i].x==point[j].x)&&(point[i].y==point[j].y)) || ((point[i].x+1==point[j].x)&&(point[i].y==point[j].y)) ||((point[i].x+1==point[j].x)&&(point[i].y==point[j].y+1)) || ((point[i].x==point[j].x)&&(point[i].y==point[j].y+1))  )

                    {

                          ma[point[i].num][point[j].num]=1;

                    }

             }

         }

         fin();

     }

 }