题目链接：http://poj.org/problem?id=3041
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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 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). Source |
这也是一道比较简单的二分匹配的题目。把光束当作图的顶点，而把小行星当作连接对应光束的边，如此一来，光束的攻击方案即对应一个顶点集合s。图中每一条边至少有一个属于s的端点。转换成了最小顶点覆盖问题。
在二分图中 最大匹配=最小顶点覆盖
于是用二分匹配可解：
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int MAX=1010;
vector<int> G[MAX];
int match[MAX],r[MAX],c[MAX];
bool used[MAX];
int k,V,n;
void add_edge(int u,int v)
{
G[u].push_back(v);
G[v].push_back(u);
}
bool dfs(int v)
{
used[v]=true;
for(int i=0;i<G[v].size();i++)
{
int u=G[v][i],w=match[u];
if(w<0||!used[w]&&dfs(w))
{
match[v]=u;
match[u]=v;
return true;
}
}
return false;
}
int bipartite_matching()
{
int res=0;
memset(match,-1,sizeof(match));
for(int v=1;v<=V;v++)
{
if(match[v]<0)
{
memset(used,0,sizeof(used));
if(dfs(v))
{
res++;
}
}
}
return res;
}
int main()
{
while(cin>>n>>k)
{
V=n*2;
for(int i=0;i<k;i++)
{
int r,c;
scanf("%d%d",&r,&c);
add_edge(r,n+c);
}
cout<<bipartite_matching()<<endl;
}
return 0;
}
poj3041(二分匹配简单题)