ZOJ 3156 Taxi (二分 + 二分匹配)

题意：给定 n 个人坐标， m 辆车的坐标，还有人的速度，要求每个人要进一辆不同的车，问你所有都进车的最短时间是多少。

析：首先二分时间 mid，很明显就是最后那个人进车的时间，然后如果把第 i 个人到时第 j 辆车的时间小于 mid，那么就从 i 向 j + n 连一条边，然后进行十分匹配，如果是完全匹配，匹配数等于 n，那么就是可以的，否则就是不可以。

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>
#include <numeric>
#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 be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int maxn = 100 + 10;
const int maxm = 1e6 + 10;
const LL mod = 1000000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -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;
}
inline int readInt(){ int x;  scanf("%d", &x);  return x; }

int x[maxn], y[maxn];
int a[maxn], b[maxn];
double v;
LL d[maxn][maxn];

LL dist(int i, int j){
  return sqr((LL)(a[i]-x[j])) + sqr((LL)(b[i]-y[j]));
}

int match[maxn<<1];
bool vis[maxn<<1];
vector<int> G[maxn<<1];

void addEdge(int u, int v){
  G[u].pb(v);  G[v].pb(u);
}

void init(){
  FOR(i, n+m+5, 0)  G[i].cl;
}

bool dfs(int u){
  vis[u] = 1;
  for(int i = 0; i < G[u].sz; ++i){
    int v = G[u][i];
    int w = match[v];
    if(w < 0 || !vis[w] && dfs(w)){
      match[u] = v;
      match[v] = u;
      return true;
    }
  }
  return false;
}

bool judge(double mid){
  int ans = 0;
  init();
  FOR(i, n, 0)  FOR(j, m, 0)
    if((double)d[i][j] <= sqr(mid * v))
      addEdge(i, j + n);
  ms(match, -1);
  for(int i = 0; i < n; ++i)  if(match[i] < 0){
    ms(vis, 0);  if(dfs(i))  ++ans;
  }
  return ans == n;
}

int main(){
  while(scanf("%d %d", &n, &m) == 2){
    for(int i = 0; i < n; ++i)  scanf("%d %d", a + i, b + i);
    for(int i = 0; i < m; ++i){
      scanf("%d %d", x + i, y + i);
      for(int j = 0; j < n; ++j)
        d[j][i] = dist(j, i);
    }
    scanf("%lf", &v);
    double l = 0., r = 1e7 / v;
    while(r-l > eps){
      double mid = (l + r) / 2.;
      if(judge(mid))  r = mid;
      else l = mid;
    }
    printf("%.2f\n", l);
  }
  return 0;
}