POJ3565

题目大意：

给定\(n\)个蚂蚁和\(n\)颗苹果树的坐标，要求每个蚂蚁爬到一颗苹果树旁，使得每个蚂蚁路线不相交且路线总长度最小，求每个蚂蚁爬到哪个苹果树旁？

首先假设有两只蚂蚁路径相交，那么这两个蚂蚁交换目标一定使得总路线缩短且不相交，所以总长度最短时所有蚂蚁路线一定不相交

怎么让总路线最短呢？二分图最小权匹配

其实只要把边权全部取反然后跑最大权匹配就好了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
namespace red{
#define int long long
#define eps (1e-6)
	inline int read()
	{
		int x=0;char ch,f=1;
		for(ch=getchar();(ch<'0'||ch>'9')&&ch!='-';ch=getchar());
		if(ch=='-') f=0,ch=getchar();
		while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
		return f?x:-x;
	}
	const int N=210;
	int n;
	struct node
	{
		int x,y;
	}ant[N],tr[N];
	double jx[N][N];
	inline int sqr(int x){return x*x;}
	inline double dis(int i,int j)
	{
		return sqrt(sqr(ant[i].x-tr[j].x)+sqr(ant[i].y-tr[j].y));
	}
	double exl[N],exr[N],slack[N];
	bool visl[N],visr[N];
	int f[N],g[N];
	inline bool find(int x)
	{
		visl[x]=1;
		for(int y=1;y<=n;++y)
		{
			if(visr[y]) continue;
			double tmp=exl[x]+exr[y]-jx[x][y];
			if(fabs(tmp)<eps)
			{
				visr[y]=1;
				if(!f[y]||find(f[y]))
				{
					f[y]=x;
					g[x]=y;
					return 1;
				}
			}
			else slack[y]=min(tmp,slack[y]);
		}
		return 0;
	}
	inline void km()
	{
		for(int i=1;i<=n;++i)
		{
			exl[i]=jx[i][1];
			for(int j=2;j<=n;++j)
			{
				exl[i]=max(exl[i],jx[i][j]);
			}
		}
		for(int i=1;i<=n;++i)
		{
			for(int j=1;j<=n;++j)
			{
				visl[j]=visr[j]=0;
				slack[j]=1e9+7;
			}
			if(find(i)) continue;
			while("haku")
			{
				double tmp=1e9+7;
				int t;
				for(int j=1;j<=n;++j)
					if(!visr[j]) tmp=min(tmp,slack[j]);
				for(int j=1;j<=n;++j)
				{
					if(visl[j]) exl[j]-=tmp;
					if(visr[j]) exr[j]+=tmp;
					else
					{
						slack[j]-=tmp;
						if(fabs(slack[j])<eps) t=j;
					}
				}
				if(!f[t]) break;
				visr[t]=1,visl[f[t]]=1;
				t=f[t];
				for(int j=1;j<=n;++j)
					slack[j]=min(slack[j],exl[t]+exr[j]-jx[t][j]);
			}
			memset(visl,0,sizeof(visl));
			memset(visr,0,sizeof(visr));
			find(i);
		}
		for(int i=1;i<=n;++i) printf("%lld\n",g[i]);
	}
	inline void main()
	{
		while(~scanf("%lld",&n))
		{
			memset(f,0,sizeof(f));
			memset(g,0,sizeof(g));
			memset(exl,0,sizeof(exl));
			memset(exr,0,sizeof(exr));
			for(int i=1;i<=n;++i)
			{
				ant[i].x=read(),ant[i].y=read();
			}
			for(int i=1;i<=n;++i)
			{
				tr[i].x=read(),tr[i].y=read();
			}
			for(int i=1;i<=n;++i)
			{
				for(int j=1;j<=n;++j)
				{
					jx[i][j]=-dis(i,j);
				}
			}
			km();
		}

	}
}
signed main()
{
	red::main();
return 0;
}