poj3656

题解：

二分图最大匹配

根据三角形不等式

直接上最大匹配即可

注意编圈取相反数

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int N=110;
const double INF=0xffffffffffff,eps=1e-6;
struct Node{double x,y;}Dot1[N],Dot2[N];
double Dist(Node a,Node b)
{
    return -sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int n,NX,NY,link[N],visx[N],visy[N];
double Map[N][N],lx[N],ly[N],slack[N];
int FindPath(int u)
{
    visx[u]=1;
    for (int i=1;i<=NY;i++)
     {
        if (visy[i])continue;
        double temp=lx[u]+ly[i]-Map[u][i];
        if (fabs(temp)<=eps)
         {
            visy[i]=1;
            if (link[i]==-1||FindPath(link[i]))
             {
                link[i]=u;
                return 1;
             }
         }
        else if (slack[i]>temp)slack[i]=temp;
    }
    return 0;
}
void KM()
{
    memset(lx,0,sizeof(lx));
    memset(ly,0,sizeof(ly));
    memset(link,-1,sizeof(link));
    for (int i=1;i<=NX;i++)
     for (int j=1;j<=NY;j++)
      if (Map[i][j]>lx[i])lx[i]=Map[i][j];
    for (int i=1;i<=NX;i++)
     {
        for (int j=1;j<=NY;j++)slack[j]=INF;
        while (1)
         {
            memset(visx,0,sizeof(visx));
            memset(visy,0,sizeof(visy));
            if (FindPath(i))break;
            double d=INF;
            for (int j=1;j<=NY;j++)
             if (!visy[j]&&d>slack[j])d=slack[j];
            for (int j=1;j<=NX;j++)
             if (visx[j])lx[j]-=d;
            for (int j=1;j<=NY;j++)
             {
                if (visy[j])ly[j]+=d;
                else slack[j]-=d;
             }
         }
     }
}
int main()
{
    while (~scanf("%d",&n))
     {
        memset(Map,0,sizeof(Map));
        for (int i=1;i<=n;i++)scanf("%lf%lf",&Dot1[i].x,&Dot1[i].y);
        for (int i=1;i<=n;i++)scanf("%lf%lf",&Dot2[i].x,&Dot2[i].y);
        for (int i=1;i<=n;i++)
         for (int j=1;j<=n;j++)Map[i][j]=Dist(Dot1[i],Dot2[j]);
        NX=NY=n;
        KM();
        for (int i=1;i<=N;i++)
         for (int j=1;j<=N;j++)
          if (link[j]==i)
           {
            printf("%d\n",j);
            break;
           }
     }
    return 0;
}