zoukankan      html  css  js  c++  java
  • UVaLive 3645 Objective: Berlin (最大流)

    题意:有n个城市,m条航班。已知每条航班的起点和终点,还有每条航班的载客量、出发时间、到达时间。并且要求在任何一个城市(起点、终点除外)都至少要有30分钟的中转时间,求起点到终点的最大客流量。

    析:把每个航线看成一个点,然后拆成两个点,然后如果两个航线能够到达,并且时间不超的话,就连一条边,然后加一个源点和汇点,分别向开始和结束城市进行连线,容量无限大。

    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 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 LL LNF = 1e17;
    const double inf = 1e20;
    const double PI = acos(-1.0);
    const double eps = 1e-3;
    const int maxn = 1e4 + 10;
    const int maxm = 1e7 + 10;
    const int mod = 1000000007;
    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;
    }
    
    struct Edge{
      int from, to, cap, flow;
    };
    struct Dinic{
      int n, m, s, t;
      vector<Edge> edges;
      vector<int> G[maxn];
      int d[maxn];
      bool vis[maxn];
      int cur[maxn];
    
      void init(int n){
        this-> n = n;
        for(int i = 0; i < n; ++i)  G[i].cl;
        edges.cl;
      }
    
      void addEdge(int from, int to, int cap){
        edges.pb((Edge){from, to, cap, 0});
        edges.pb((Edge){to, from, 0, 0});
        m = edges.sz;
        G[from].pb(m - 2);
        G[to].pb(m - 1);
      }
    
      bool bfs(){
        ms(vis, 0);  vis[s] = 1;  d[s] = 0;
        queue<int> q;  q.push(s);
    
        while(!q.empty()){
          int u = q.front();  q.pop();
          for(int i = 0; i < G[u].sz; ++i){
            Edge &e = edges[G[u][i]];
            if(!vis[e.to] && e.cap > e.flow){
              vis[e.to] = 1;
              d[e.to] = d[u] + 1;
              q.push(e.to);
            }
          }
        }
        return vis[t];
      }
    
      int dfs(int u, int a){
        if(u == t || a == 0)  return a;
        int flow = 0, f;
        for(int &i = cur[u]; i < G[u].sz; ++i){
          Edge &e = edges[G[u][i]];
          if(d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){
            e.flow += f;
            edges[G[u][i]^1].flow -= f;
            flow += f;
            a -= f;
            if(a == 0)  break;
          }
        }
        return flow;
      }
    
      int maxflow(int s, int t){
        this-> s = s;
        this-> t = t;
        int flow = 0;
        while(bfs()){ ms(cur, 0);  flow += dfs(s, INF); }
        return flow;
      }
    };
    
    Dinic dinic;
    map<string, int> mp;
    
    int ID(const string &s){
      if(mp.count(s))  return mp[s];
      return mp[s] = mp.sz;
    }
    
    struct Fly{
      int s, t, st, tt;
    };
    Fly fly[maxn];
    
    int main(){
      ios::sync_with_stdio(false);
      while(cin >> n){
        mp.cl;
        int s, t, last, num;
        string city1, city2;
        cin >> city1;  s = ID(city1);
        cin >> city2;  t = ID(city2);
        cin >> last;
        last = last / 100 * 60 + last % 100;
        cin >> m;
        int S = m<<1, T = m<<1|1;
        dinic.init(T + 3);
        for(int i = 0; i < m; ++i){
          cin >> city1 >> city2 >> num >> fly[i].st >> fly[i].tt;
          fly[i].s = ID(city1);
          fly[i].t = ID(city2);
          fly[i].st = fly[i].st / 100 * 60 + fly[i].st % 100;
          fly[i].tt = fly[i].tt / 100 * 60 + fly[i].tt % 100;
          if(fly[i].s == s)  dinic.addEdge(S, i<<1, INF);
          if(fly[i].t == t && fly[i].tt <= last)  dinic.addEdge(i<<1|1, T, INF);
          dinic.addEdge(i<<1, i<<1|1, num);
          for(int j = 0; j < i; ++j){
            if(fly[i].t == fly[j].s && fly[i].tt + 30 <= fly[j].st)  dinic.addEdge(i<<1|1, j<<1, INF);
            else if(fly[j].t == fly[i].s && fly[j].tt + 30 <= fly[i].st)  dinic.addEdge(j<<1|1, i<<1, INF);
          }
        }
        cout << dinic.maxflow(S, T) << endl;
      }
      return 0;
    }
    

      

  • 相关阅读:
    深入理解Auto Layout 第一弹
    Android UI测量、布局、绘制过程探究
    Android Measure 体系简单总结
    [译]Android view 测量布局和绘制的流程
    UI布局只关乎三件事情:尺寸、位置、组织
    ios 布局 素材 待整理
    NSAllowsArbitraryLoadsInWebContent NSAllowsArbitraryLoads
    UIStackView
    laravel 自定义常量方法
    laravel 增加不存在数据库的字段
  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7761385.html
Copyright © 2011-2022 走看看