[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\n", ans);
}

Excited!

posted on 2017-02-22 15:07 蒟蒻konjac 阅读(...) 评论(...) 编辑 收藏

统计