poj3041——最小点覆盖

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).

 

 

 

大意：K个目标在N*N的方阵中，一次攻击可以消除同一行或同一列的目标，问最少几次攻击消除所有目标。

题解：这题是看题解写的，建一个二分图，图一侧的定点代表行，另一侧的定点代表列，如果（x，y）有目标，就连边x->y。

那么这题就转换为最少选多少个点，使所有边的两个定点至少有一个被选中，这就是经典模型最小点覆盖。

结论：最小点覆盖=最大匹配。

证明：详见http://blog.sina.com.cn/s/blog_5ceeb9ea0100l08n.html

写的很好

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<string>
using namespace std;
int read(){
    int xx=0,ff=1;char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')ff=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){xx=(xx<<3)+(xx<<1)+ch-'0';ch=getchar();}
    return xx*ff;
}
const int maxn=1010;
int N,K,lin[maxn],len;
struct edge{
	int y,next;
}e[10010];
inline void insert(int xx,int yy){
	e[++len].next=lin[xx];
	lin[xx]=len;
	e[len].y=yy;
}
int tim,pretim,match[maxn],vis[maxn],ans;
bool hun(int x){
	for(int i=lin[x];i;i=e[i].next)
		if(vis[e[i].y]<=pretim){
			vis[e[i].y]=++tim;
			if(match[e[i].y]==0||hun(match[e[i].y])){
				match[x]=e[i].y;
				match[e[i].y]=x;
				return 1;
			}
		}
	return 0;
}
int main(){
	//freopen("in","r",stdin);
	//freopen("out","w",stdout);
	N=read(),K=read();
	for(int i=1;i<=K;i++){
		int t1=read(),t2=read()+N;
		insert(t1,t2);
	}
	for(int i=1;i<=N*2;i++)
		if(!match[i]){
			pretim=tim;
			tim++;
			if(hun(i))
				ans++;
		}
	printf("%d\n",ans);
	return 0;
}