UVaLive 4043 Ants (最佳完美匹配)

题意：给定 n 个只蚂蚁和 n 棵树的坐标，问怎么匹配使得每个蚂蚁到树的连线不相交。

析：可以把蚂蚁和树分别看成是两类，那么就是一个完全匹配就好，但是要他们的连线不相交，那么就得考虑，最佳完美匹配是可以的，为什么呢，假设有两条线段a1-b1和a2-b2，那么如果相交，dist(a1, b1) + dist(a2, b2) > dist(a1, b2) + dist(a2, b1)，自己画图看看就好，然后如果是最小的距离，那么就是可以的了，只要在用使得KM匹配时，把权值取反就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 50;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

double w[maxn][maxn], x[maxn], y[maxn];
int prex[maxn], prey[maxn], son[maxn];
double slack[maxn];
int par[maxn];

void adjust(int v){
  son[v] = prey[v];
  if(prex[son[v]] != -2)  adjust(prex[son[v]]);
}

bool Find(int v){
  FOR(i, 0, n) if(prey[i] == -1){
    if(slack[i] > x[v] + y[i] - w[v][i]){
      slack[i] = x[v] + y[i] - w[v][i];
      par[i] = v;
    }
    if(x[v] + y[i] == w[v][i]){
      prey[i] = v;
      if(son[i] == -1){
        adjust(i);
        return true;
      }
      if(prex[son[i]] != -1)  continue;
      prex[son[i]] = i;
      if(Find(son[i]))  return true;
    }
  }
  return false;
}

int KM(){
  ms(son, -1);  ms(y, 0);
  for(int i = 0; i < n; ++i){
    x[i] = 0;
    for(int j = 0; j < n; ++j)
      x[i] = max(x[i], w[i][j]);
  }
  bool ok;
  for(int i = 0; i < n; ++i){
    for(int j = 0; j < n; ++j){
      prex[j] = prey[j] = -1;
      slack[j] = inf;
    }
    prex[i] = -2;
    if(Find(i))  continue;
    ok = false;
    while(!ok){
      double m = inf;
      FOR(j, 0, n)  if(prey[j] == -1)  m = min(m, slack[j]);
      for(int j = 0; j < n; ++j){
        if(prex[j] != -1)  x[j] -= m;
        if(prey[j] != -1)  y[j] += m;
        else slack[j] -= m;
      }
      FOR(j, 0, n)  if(prey[j] == -1 && slack[j] == 0.0){
        prey[j] = par[j];
        if(son[j] == -1){
          adjust(j);
          ok = true;
          break;
        }
        prex[son[j]] = j;
        if(Find(son[j])){
          ok = true;
          break;
        }
      }
    }
  }
}

double a[maxn], b[maxn];

int main(){
  while(scanf("%d", &n) == 1){
    for(int i = 0; i < n; ++i)  scanf("%lf %lf", a + i, b + i);
    for(int i = 0; i < n; ++i){
      double x, y;
      scanf("%lf %lf", &x, &y);
      for(int j = 0; j < n; ++j)  w[i][j] = -sqrt(sqr(x-a[j]) + sqr(y-b[j]));
    }
    KM();
    for(int i = 0; i < n; ++i)
      printf("%d
", son[i] + 1);
  }
  return 0;
}