UVALive 4043 Ants

KM   构图求最小权值匹配

保证最小的权值,所连的边一定是能够不相交的.

[Submit]   [Go Back]   [Status]  

Description

Young naturalist Bill studies ants in school. His ants feed on plant-louses that live on apple trees. Each ant colony needs its own apple tree to feed itself.

Bill has a map with coordinates of n ant colonies and n apple trees. He knows that ants travel from their colony to their feeding places and back using chemically tagged routes. The routes cannot intersect each other or ants will get confused and get to the wrong colony or tree, thus spurring a war between colonies.

Bill would like to connect each ant colony to a single apple tree so that all n routes are non-intersecting straight lines. In this problem such connection is always possible. Your task is to write a program that finds such connection.

On this picture ant colonies are denoted by empty circles and apple trees are denoted by filled circles. One possible connection is denoted by lines.

5 
-42 58 
44 86 
7 28 
99 34 
-13 -59 
-47 -44 
86 74 
68 -75 
-68 60 
99 -60

4 
2 
1 
5 
3

Source

[Submit]   [Go Back]   [Status]  

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

const int maxn=220;
const double INF=999999999999999.;
const double eps=1e-5;

struct Dian
{
    double x,y;
}white[maxn*maxn],black[maxn*maxn];

int n;
double g[maxn][maxn];
int linker[maxn*maxn];
double lx[maxn*maxn],ly[maxn*maxn];
double slack[maxn*maxn];
bool visx[maxn*maxn],visy[maxn*maxn];

bool dfs(int x)
{
    visx[x]=true;
    for(int y=0;y<n;y++)
    {
        if(visy[y]) continue;
        double tmp=lx[x]+ly[y]-g[x][y];
        if(fabs(tmp)<eps)
        {
            visy[y]=true;
            if(linker[y]==-1||dfs(linker[y]))
            {
                linker[y]=x;
                return true;
            }
        }
        else if(slack[y]>tmp)
            slack[y]=tmp;
    }
    return false;
}

int KM()
{
    memset(linker,-1,sizeof(linker));
    memset(ly,0,sizeof(ly));
    for(int i=0;i<n;i++)
    {
        lx[i]=-INF;
        for(int j=0;j<n;j++)
            if(g[i][j]>lx[i])
                lx[i]=g[i][j];
    }
    for(int x=0;x<n;x++)
    {
        for(int i=0;i<n;i++)
            slack[i]=INF;
        while(true)
        {
            memset(visx,false,sizeof(visx));
            memset(visy,false,sizeof(visy));
            if(dfs(x)) break;
            double d=INF;
            for(int i=0;i<n;i++)
                if(!visy[i]&&d>slack[i])
                    d=slack[i];
            for(int i=0;i<n;i++)
                if(visx[i])
                    lx[i]-=d;
            for(int i=0;i<n;i++)
                if(visy[i])
                    ly[i]+=d;
                else
                    slack[i]-=d;
        }
    }
}

double Dist(Dian a,Dian b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

int main()
{
    bool first=false;
    while(scanf("%d",&n)!=EOF)
    {
        if(first) putchar(10);
        else first=true;
        for(int i=0;i<n;i++)
        {
            double a,b;
            scanf("%lf%lf",&a,&b);
            white[i]=(Dian){a,b};
        }
        for(int i=0;i<n;i++)
        {
            double a,b;
            scanf("%lf%lf",&a,&b);
            black[i]=(Dian){a,b};
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                double t=Dist(black[i],white[j]);
                g[i][j]=-t;
            }
        }
        KM();
        for(int i=0;i<n;i++)
        {
            printf("%d\n",linker[i]+1);
        }
    }
    return 0;
}