POJ3041--Asteroids--二分图最大匹配--Konig

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.

//此题用到Konig定理。即最少覆盖数==最大匹配数。

然后就是个很简单的二分图最大匹配问题。

X集为横坐标，Y集为纵坐标。xiyi边为炸弹。则只需要每条边（即每个炸弹）至少有一个点和xi或者yi关联。然后就是Konig

#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
vector <int> ans[1008];
bool vis[1008];
int marry[1008];
int sum;
bool dfs(int u)
{
	for(int i=0;i<ans[u].size();i++)
	{
		if(!vis[ans[u][i]])
		{
			vis[ans[u][i]]=1;
			if(!marry[ans[u][i]]||dfs(marry[ans[u][i]]))
			{
				marry[ans[u][i]]=u;//这里不用互为夫妻。。因为dfs的永远是一方。
				return 1;
			}
		}
	}
	return 0;
}
int main()
{
	int n,k;//n<=500,k<=10000
	while(scanf("%d%d",&n,&k)==2)
	{
		sum=0;
		int u,v;
		for(int i=1;i<=2*n+10;i++)
		{
			ans[i].clear();
		}
		for(int i=1;i<=k;i++)
		{
			scanf("%d%d",&u,&v);
			v=v+n;
			ans[u].push_back(v);
		}
		for(int i=1;i<=n;i++)
		{
			memset(vis,0,sizeof(vis));
			if(dfs(i))
			{
				sum++;
			}
		}
		cout<<sum<<endl;
	}
	return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 1008
#define edge 21000
#define inf 0x3f3f3f3f
int first[maxn],dis[maxn],num[maxn];
int vv[edge],ww[edge],nxt[edge];
int e,NN;
void addEdge(int u,int v,int w)
{
	vv[e] = v;ww[e] = w;nxt[e] = first[u];first[u] = e++;
	vv[e] = u;ww[e] = 0;nxt[e] = first[v];first[v] = e++;
}
inline int min(int a,int b)
{
	return a>b?b:a;
}
int dfs(int u,int s,int d,int cost)
{
	if(u == d) return cost;
	int ans = 0;
	int _min = NN;
	for(int i=first[u];i!=-1;i=nxt[i])
	{
		int v = vv[i];
		if(ww[i])
		{
			if(dis[v] + 1 == dis[u])
			{
				int t = dfs(v,s,d,min(ww[i],cost));
				ww[i] -= t;
				ww[i^1] += t;
				ans += t;
				cost -= t;
				if(dis[s] == NN) return ans;
				if(!cost) break;
			}
			if(_min > dis[v])
					_min = dis[v];
		}
	}
	if(!ans)
	{
		if(--num[dis[u]] == 0) dis[s] = NN;
		dis[u] = _min + 1;
		++num[dis[u]];
	}
	return ans;
}
int isap(int s,int d)
{
	memset(dis,0,sizeof(dis));
	memset(num,0,sizeof(num));
	num[0] = NN;
	int ans = 0;
	while(dis[s] < NN)
			ans += dfs(s,s,d,inf);
	return ans;
}
int main()
{
	//freopen("in.txt","r",stdin);
	int n,m;
	while(scanf("%d%d",&n,&m)==2)
	{
		memset(first,-1,sizeof(first));
		e = 0;
		NN = 2*n + 2;
		for(int i=1;i<=n;i++)
		{
			addEdge(0,i,1);
		}
		for(int i=n+1;i<=2*n;i++)
		{
			addEdge(i,2*n+1,1);
		}
		for(int i=1;i<=m;i++)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			v += n;
			addEdge(u,v,1);
		}
		int ans = isap(0,2*n+1);
		printf("%d\n",ans);
	}
	return 0;
}