• UVa 753 A Plug for UNIX (最大流)


    题意:给定 n 种插座,m种设备,和k个转换器,问你最少有几台设备不能匹配。

    析:一个很裸的网络流,直接上模板就行,建立一个源点s和汇点t,源点和每个设备连一条边,每个插座和汇点连一条边,然后再连转换器,

    最后跑一次最大流即可。

    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>
    #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 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 = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-5;
    const int maxn = 500 + 10;
    const int mod = 1e6;
    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];
      bool vis[maxn];
      int d[maxn];
      int cur[maxn];
    
      void init(){
        edges.clear();
        for(int i = 0; i < maxn; ++i)  G[i].clear();
      }
      bool bfs(){
        memset(vis, 0, sizeof vis);
        queue<int> q;
        q.push(s);
        d[s] = 0;  vis[s] = true;
        while(!q.empty()){
          int x = q.front();  q.pop();
          for(int i = 0; i < G[x].size(); ++i){
            Edge &e = edges[G[x][i]];
            if(!vis[e.to] && e.cap > e.flow){
              vis[e.to] = 1;
              d[e.to] = d[x] + 1;
              q.push(e.to);
            }
          }
        }
        return vis[t];
      }
    
      int dfs(int x, int a){
        if(x == t || a == 0)  return a;
        int flow = 0, f;
        for(int& i = cur[x]; i < G[x].size(); ++i){
          Edge& e = edges[G[x][i]];
          if(d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){
            e.flow += f;
            edges[G[x][i]^1].flow -= f;
            flow += f;
            a -= f;
            if(!a)  break;
          }
        }
        return flow;
      }
    
      int maxflow(int s, int t){
        this->s = s;  this->t = t;
        int flow = 0;
        while(bfs()){
          memset(cur, 0, sizeof cur);
          flow += dfs(s, INF);
        }
        return flow;
      }
    
      void addEdge(int from, int to, int cap){
        edges.push_back(Edge{from, to, cap, 0});
        edges.push_back(Edge{to, from, 0, 0});
        m = edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
      }
    };
    map<string, int> mp;
    
    int getId(const string &s){
      if(mp.count(s))  return mp[s];
      return mp[s] = mp.size();
    }
    vector<int> socket;
    vector<P> device;
    vector<P> convert;
    
    void init(){
      mp.clear();
      socket.clear();
      device.clear();
      convert.clear();
    }
    
    int solve(){
      int s = mp.size() + 1;
      int t = mp.size() + 2;
      Dinic dinic;
      dinic.init();
      for(int i = 0; i < device.size(); ++i){
        dinic.addEdge(s, device[i].first, 1);
        dinic.addEdge(device[i].first, device[i].second, 1);
      }
      for(int i = 0; i < socket.size(); ++i)
        dinic.addEdge(socket[i], t, 1);
      for(int i = 0; i < convert.size(); ++i)
        dinic.addEdge(convert[i].first, convert[i].second, INF);
      return (int)device.size() -dinic.maxflow(s, t);
    }
    
    int main(){
      int T;  cin >> T;
      while(T--){
        cin >> n;
        init();
        string s;
        for(int i = 0; i < n; ++i){
          cin >> s;
          socket.push_back(getId(s));
        }
        cin >> n;
        string ss;
        for(int i = 0; i < n; ++i){
          cin >> s >> ss;
          device.push_back(P(getId(s), getId(ss)));
        }
        cin >> n;
        for(int i = 0; i < n; ++i){
          cin >> s >> ss;
          convert.push_back(P(getId(s), getId(ss)));
        }
        cout << solve() << endl;
        if(T) cout << endl;
      }
      return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/6415907.html
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