Ants(Northeastern Europe 2007)

Ants

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 3539		Accepted: 1064		Special Judge

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 ofn ant colonies andn 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 alln 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.

Input

The first line of the input file contains a single integer numbern (1 ≤n ≤ 100) — the number of ant colonies and apple trees. It is followed by n lines describing n ant colonies, followed by n lines describing n apple trees. Each ant colony and apple tree is described by a pair of integer coordinatesx andy (−<nobr>10 000</nobr> ≤x,y ≤<nobr>10 000</nobr>) on a Cartesian plane. All ant colonies and apple trees occupy distinct points on a plane. No three points are on the same line.

Output

Write to the output filen lines with one integer number on each line. The number written oni-th line denotes the number (from 1 ton) of the apple tree that is connected to thei-th ant colony.

Sample Input

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

Sample Output

4
2
1
5
3

Source

Northeastern Europe 2007

在暑假集训结束的最后两天再学习一个新的算法，来满足一下自己的成就感。。哈哈！！！

但是有好多不理解是为什么，希望理解的大牛，大神可以解释一下。

题意：

题目的意思就是说有一个蚂蚁军团，想爬上苹果树吃苹果。然后题目会先给你一个n代表蚂蚁军团和苹果树各有n个。然后叫你通过计算而确定如何连接。

从题目中可以看到，苹果树和蚂蚁分别分成了两个部分。因此，可以轻松的想到了二分匹配问题。

然后就可以根据数学几何的知识，如果a1--b1与a2--b2相交，那么dist(a1,b1)+dist(a2,b2)一定大于dist(a1,b2)+dist(a2,b1),因为三角形的两边之和大于第三边。因此可得知最佳匹配中不会出现线段相交的情况。

还要注意输出的时候，因为是树按照从1--n输出的所以要进行转换后再输出。

题目意思如果知道了会，但不知细节问题的话，那就看下面的两个代码吧。

一个是O（n^4）的KM( ),（还有一个是经过松弛处理的KM（）；不过非本人原创）。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#define CL(x,v);memset(x,v,sizeof(x));
#define maxn 105
#define INF 1000000
using namespace std;

int Left[maxn],n;
bool S[maxn],T[maxn];
double Lx[maxn],Ly[maxn],w[maxn][maxn];
struct Point
{
    double x,y;
}p[maxn*2];

double Dist(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
bool match(int i)
{
    S[i] = true;
    for(int j = 1;j <= n;j++)
      if(fabs(Lx[i]+Ly[j]-w[i][j])<1e-5&&!T[j])
      {
          T[j] = true;
          if(!Left[j]||match(Left[j])){
             Left[j] = i;
             return true;
          }
      }
    return false;
}
 //    没有这个函数，则就是稳定婚姻模型了。
void update()
{
    double a = INF;
    for(int i = 1;i <= n;i++) if(S[i])
      for(int j = 1;j <= n;j++) if(!T[j])
        a = min(a,Lx[i]+Ly[j]-w[i][j]);
    for(int i = 1;i <= n;i++){
      if(S[i]) Lx[i] -= a;
      if(T[i]) Ly[i] += a;
    }
}

void KM()
{
    CL(Left,0);CL(Lx,0);CL(Ly,0);
    for(int i = 1;i <= n;i++)
      for(int j = 1;j <= n;j++)
        Lx[i] = max(Lx[i],w[i][j]);
    for(int i = 1;i <= n;i++){
       while(1){
         CL(S,0); CL(T,0);
         if(match(i)) break;
         else update();
       }
    }
}
int main()
{
    while(cin>>n)
    {
        for(int i = 1;i <= 2*n;i++)
          cin>>p[i].x>>p[i].y;
        for(int i = 1;i <= n;i++)
          for(int j = 1;j <= n;j++)
            w[i][j] = -Dist(p[i],p[j+n]);  //这里为嘛？加一个负号。
        KM();                              //哪位大牛，大神知道的请告诉一下哈
        int tree[maxn];
        for(int i = 1;i <= n;i++)
          tree[Left[i]] = i;
        for(int i = 1;i <= n;i++)
          cout<<tree[i]<<endl;
    }
    return 0;
}