Uva 1411 - Ants（KM模板O(n^4)）最大权匹配

连接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=560&problem=4157&mosmsg=Submission+received+with+ID+11685541

 

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 <tex2html_verbatim_mark>ant colonies and n <tex2html_verbatim_mark>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 <tex2html_verbatim_mark>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.

 

<tex2html_verbatim_mark>

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 

Input has several dataset. The first line of each dataset contains a single integer number n <tex2html_verbatim_mark>(1n100) <tex2html_verbatim_mark>-- the number of ant colonies and apple trees. It is followed by n <tex2html_verbatim_mark>lines describing n <tex2html_verbatim_mark>ant colonies, followed byn <tex2html_verbatim_mark>lines describing n <tex2html_verbatim_mark>apple trees. Each ant colony and apple tree is described by a pair of integer coordinates x <tex2html_verbatim_mark>and y <tex2html_verbatim_mark>(- 10000x, y10000) <tex2html_verbatim_mark>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 

For each dataset, write to the output file n <tex2html_verbatim_mark>lines with one integer number on each line. The number written on i <tex2html_verbatim_mark>-th line denotes the number (from 1 to n <tex2html_verbatim_mark>) of the apple tree that is connected to the i i<tex2html_verbatim_mark>-th ant colony. Print a blank line between datasets.

 

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

 

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int maxn = 100 + 10;

const double INF = 1e30;

int n;
double W[maxn][maxn];
double Lx[maxn], Ly[maxn];   // 顶标
int left[maxn];          // left[i]为右边第i个点的匹配点编号
bool S[maxn], T[maxn];   // S[i]和T[i]为左/右第i个点是否已标记

bool eq(double a, double b) {
  return fabs(a-b) < 1e-9;
}

bool match(int i){
  S[i] = true;
  for(int j = 1; j <= n; j++) if (eq(Lx[i]+Ly[j], W[i][j]) && !T[j]){
    T[j] = true;
    if (!left[j] || match(left[j])){
      left[j] = i;
      return true;
    }
  }
  return false;
}

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() {
  for(int i = 1; i <= n; i++) {
    left[i] = Lx[i] = Ly[i] = 0;
    for(int j = 1; j <= n; j++)
      Lx[i] = max(Lx[i], W[i][j]);
  }
  for(int i = 1; i <= n; i++) {
    for(;;) {
      for(int j = 1; j <= n; j++) S[j] = T[j] = 0;
      if(match(i)) break; else update();
    }
  }
}

int main(){
  int kase = 0;
  while(scanf("%d", &n) == 1) {
    if(++kase > 1) printf("\n");

    int x1[maxn], y1[maxn], x2[maxn], y2[maxn];
    for(int i = 1; i <= n; i++) scanf("%d%d", &x1[i], &y1[i]);
    for(int i = 1; i <= n; i++) scanf("%d%d", &x2[i], &y2[i]);
    for(int i = 1; i <= n; i++) // ant colony
      for(int j = 1; j <= n; j++) //存成负数找最大值，相当于找最小值
        W[j][i] = -sqrt((double)(x1[i]-x2[j])*(x1[i]-x2[j]) + (double)(y1[i]-y2[j])*(y1[i]-y2[j]));
    KM(); // 最大权匹配
    for(int i = 1; i <= n; i++) printf("%d\n", left[i]);//x1[i]这个蚂蚁与x2[left[i]]这个蚂蚁匹配
}
  return 0;
}