• [bzoj3931][CQOI2015]网络吞吐量——最短路+网络流


    题目

    传送门

    题解

    第一次一遍就AC一道bzoj上的题,虽然是一道水题。。。
    我们做一边最短路,求出每个点的dist,然后再做一次类似spfa的操作,求出每个点是否可以用于建图。
    在新图上拆点跑一边dinic就好辣。

    代码

    #include <algorithm>
    #include <cstdio>
    #include <cstring>
    #include <queue>
    #include <vector>
    using namespace std;
    #define ll long long
    const ll inf = 100000000000;
    const int maxn = 1000;
    struct edge {
      int from;
      int to;
      ll value;
    };
    vector<edge> edges;
    vector<int> G[maxn];
    vector<edge> gg[maxn];
    void add(int from, int to, ll value) {
      gg[from].push_back((edge){from, to, value});
    }
    void add_edge(int from, int to, ll value) {
      edges.push_back((edge){from, to, value});
      edges.push_back((edge){to, from, 0});
      int m = edges.size();
      G[from].push_back(m - 2);
      G[to].push_back(m - 1);
    }
    ll dist[maxn];
    int iter[maxn];
    int n, m;
    ll c[maxn];
    int st, t;
    void spfa(int s) {
      for (int i = 1; i <= n; i++)
        dist[i] = inf;
      queue<int> q;
      bool inq[maxn];
      memset(inq, 0, sizeof(inq));
      q.push(s);
      dist[s] = 0;
      inq[s] = 1;
      while (!q.empty()) {
        int u = q.front();
        q.pop();
        inq[u] = 0;
        for (int i = 0; i < gg[u].size(); i++) {
          edge &e = gg[u][i];
          if (dist[e.to] > dist[u] + e.value) {
            dist[e.to] = dist[u] + e.value;
            if (!inq[e.to]) {
              q.push(e.to);
              inq[e.to] = 1;
            }
          }
        }
      }
    }
    void build_graph(int s, int e) {
      queue<int> q;
      q.push(s);
      bool inq[maxn];
      memset(inq, 0, sizeof(inq));
      inq[s] = 1;
      int vis[maxn];
      memset(vis, 0, sizeof(vis));
      vis[s] = 1;
      add_edge(st, s, inf);
      add_edge(e + n, t, inf);
      add_edge(s, s + n, inf);
      add_edge(e, e + n, inf);
      while (!q.empty()) {
        int u = q.front();
        q.pop();
        inq[u] = 0;
        for (int i = 0; i < gg[u].size(); i++) {
          edge &ee = gg[u][i];
          if (dist[ee.to] == dist[u] + ee.value) {
    
            add_edge(u + n, ee.to, inf);
            if (!vis[ee.to] && ee.to != e) {
              add_edge(ee.to, ee.to + n, c[ee.to]);
              vis[ee.to] = 1;
            }
            if (!inq[ee.to]) {
              q.push(ee.to);
              inq[ee.to] = 1;
            }
          }
        }
      }
    }
    void bfs(int s) {
      memset(dist, -1, sizeof(dist));
      dist[s] = 0;
      queue<int> q;
      q.push(s);
      while (!q.empty()) {
        int u = q.front();
        q.pop();
        for (int i = 0; i < G[u].size(); i++) {
          edge &e = edges[G[u][i]];
          if (e.value > 0 && dist[e.to] == -1) {
            dist[e.to] = dist[u] + 1;
            q.push(e.to);
          }
        }
      }
    }
    ll dfs(int s, int t, ll flow) {
      if (s == t)
        return flow;
      for (int &i = iter[s]; i < G[s].size(); i++) {
        edge &e = edges[G[s][i]];
        if (dist[e.to] > dist[s] && e.value > 0) {
          int d = dfs(e.to, t, min(flow, e.value));
          if (d > 0) {
            e.value -= d;
            edges[G[s][i] ^ 1].value += d;
            return d;
          }
        }
      }
      return 0;
    }
    ll dinic(int s, int t) {
      ll flow = 0;
      while (1) {
        bfs(s);
        if (dist[t] == -1)
          return flow;
        memset(iter, 0, sizeof(iter));
        ll d;
        while (d = dfs(s, t, inf))
          flow += d;
      }
    }
    int main() {
      //  freopen("input", "r", stdin);
      scanf("%d %d", &n, &m);
      for (int i = 1; i <= m; i++) {
        int x, y;
        ll z;
        scanf("%d %d %lld", &x, &y, &z);
        add(x, y, z);
        add(y, x, z);
      }
      for (int i = 1; i <= n; i++)
        scanf("%lld", &c[i]);
      spfa(1);
      st = 0, t = n + n + 1;
      build_graph(1, n);
    
      ll ans = dinic(st, t);
      printf("%lld
    ", ans);
    }
    

    Excited!

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  • 原文地址:https://www.cnblogs.com/gengchen/p/6429059.html
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