CUGB图论专场2：B - Asteroids 二分图：最小顶点覆盖=最大匹配数

B - Asteroids

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status

Description

Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid contains K asteroids (1 <= K <= 10,000), which are conveniently located at the lattice points of the grid. 

Fortunately, Bessie has a powerful weapon that can vaporize all the asteroids in any given row or column of the grid with a single shot.This weapon is quite expensive, so she wishes to use it sparingly.Given the location of all the asteroids in the field, find the minimum number of shots Bessie needs to fire to eliminate all of the asteroids.

Input

* Line 1: Two integers N and K, separated by a single space. 
* Lines 2..K+1: Each line contains two space-separated integers R and C (1 <= R, C <= N) denoting the row and column coordinates of an asteroid, respectively.

Output

* Line 1: The integer representing the minimum number of times Bessie must shoot.

Sample Input

3 4
1 1
1 3
2 2
3 2

Sample Output

2

Hint

INPUT DETAILS: 
The following diagram represents the data, where "X" is an asteroid and "." is empty space: 
X.X 
.X. 
.X.

OUTPUT DETAILS: 
Bessie may fire across row 1 to destroy the asteroids at (1,1) and (1,3), and then she may fire down column 2 to destroy the asteroids at (2,2) and (3,2).

题意：就是横纵坐标求出x或者y，即点覆盖边，即是求最小点覆盖。

例如：样例中的取x=1和y=2就可以灭掉所以的星星，就是二分图中左顶点集的1点和右顶点集的2点，就可以把二分图覆盖完了（横坐标x为左顶点集，纵坐标y为右顶点集。

最小点覆盖：假如选了一个点就相当于覆盖了以它为端点的所有边，你需要选择最少的点来覆盖图的所有的边。

这题看了学习资料，学习匈牙利算法就懂了。不过还要学习二分图最大匹配的König定理：最小点覆盖数 = 最大匹配数。

所以求最小顶点覆盖等于求最大匹配数，而最大匹配数用匈牙利算法可以求出……

因为要用二分图求，所以得有左顶点集和右顶点集，而从样例中可以知道x和y是互斥的，所以可以设x为左顶点集，y为右顶点集。

邻接表：0ms可以AC。

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <list>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <set>
#define PI acos(-1.0)
#define mem(a,b) memset(a,b,sizeof(a))
#define sca(a) scanf("%d",&a)
#define pri(a) printf("%d\n",a)
#define MM 10002
#define MN 505
#define INF 168430090
using namespace std;
typedef long long ll;
int n,k,m,pre[MN],vis[MN];//pre为y匹配点x，即匹配的前一结点
int cnt,head[MM]; //邻接表，邻接矩阵也可以，不过邻接表空点不用判断，省时
struct node
{
    int x,y,next;
} e[MM];
int dfs(int x)
{
    for(int y,i=head[x]; i!=-1; i=e[i].next)
    {
        y=e[i].y;
        if(!vis[y])
        {
            vis[y]=1;
            if(pre[y]==0||dfs(pre[y])) //dfs前向结点，如果为真说明找到增广路
            {
                pre[y]=x; //把匹配的变成未匹配，未匹配的变成匹配，故匹配数+1，因为未匹配数比匹配数大1
                return 1;
            }
        }
    }
    return 0;
}
void search()
{
    for(int x=1; x<=n; x++) //从左顶点集开始判断，匈牙利算法思想
    {
        mem(vis,0);
        m+=dfs(x);
    }
}
int main()
{
    cin>>n>>k;
    mem(head,-1);
    for(int x,y,i=0; i<k; i++)
    {
        scanf("%d%d",&x,&y);
        e[cnt].y=y; e[cnt].next=head[x];
        head[x]=cnt++;
    }
    search();
    cout<<m<<endl;
    return 0;
}

邻接矩阵：32ms可以AC，因为空的点也需要判断，导致了不必要的操作，所以时间会慢些。

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <list>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <set>
#define PI acos(-1.0)
#define mem(a,b) memset(a,b,sizeof(a))
#define sca(a) scanf("%d",&a)
#define pri(a) printf("%d\n",a)
#define MM 10002
#define MN 505
#define INF 168430090
using namespace std;
typedef long long ll;
int n,k,m,pre[MN],vis[MN],Map[MN][MN];
int dfs(int x)
{
    for(int y=1;y<=n;y++)
    {
        if(!vis[y]&&Map[x][y])
        {
            vis[y]=1;
            if(pre[y]==0||dfs(pre[y])) //dfs前向结点，如果为真说明找到增广路
            {
                pre[y]=x; //把匹配的变成未匹配，未匹配的变成匹配，故匹配数+1，因为未匹配数比匹配数大1
                return 1;
            }
        }
    }
    return 0;
}
void search()
{
    for(int x=1; x<=n; x++) //从左顶点集开始判断，匈牙利算法思想
    {
        mem(vis,0);
        m+=dfs(x);
    }
}
int main()
{
    cin>>n>>k;
    for(int x,y,i=0; i<k; i++)
    {
        scanf("%d%d",&x,&y);
        Map[x][y]=1;
    }
    search();
    cout<<m<<endl;
    return 0;
}