• 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
    ", l);
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/8697882.html
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