POJ3041 Asteroids

Asteroids

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).

题目大意是，一个矩阵，某些位置有小行星，有一种炸弹，一次可以炸掉一行或者一列，现在问题是需要最少用多少这样的炸弹。

我们用匈牙利算法来求最大匹配

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6
 7 const int maxn=505;
 8 const int maxm=10005;
 9 int N,K;
10 int Head[maxn],Adj[maxm],Next[maxm];
11 int match[maxn],use[maxn];
12 int c;
13
14 void AddEdge(int u,int v)
15 {
16     c++;
17     Adj[c]=v;
18     Next[c]=Head[u];
19     Head[u]=c;
20 }
21
22 bool dfs(int x)
23 {
24     for(int i=Head[x];i;i=Next[i])
25     {
26         int v=Adj[i]-N;
27         if(use[v]) continue;
28         use[v]=true;
29         if(match[v]==0||dfs(match[v]))
30         {
31             match[v]=x;
32             return true;
33         }
34     }
35     return false;
36 }
37
38 void hungarian(int &cnt)
39 {
40     for(int i=1;i<=N;i++)
41     {
42         memset(use,false,sizeof(use));
43         if(dfs(i)) cnt++;
44     }
45 }
46
47 int main()
48 {
49     freopen("poj3041.in","r",stdin);
50     freopen("poj3041.out","w",stdout);
51     scanf("%d %d",&N,&K);
52     int a,b;
53     for(int i=1;i<=K;i++)
54     {
55         scanf("%d %d",&a,&b);
56         AddEdge(a,b+N);
57     }
58     int cnt=0;
59     hungarian(cnt);
60     printf("%d",cnt);
61     return 0;
62 }