ZOJ 3156 Taxi (二分 + 二分匹配)

时间：2022-12-24 19:10:40

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;

}

　　

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