【BZOJ2561】最小生成树

【题目链接】

• 点击打开链接

【思路要点】

• 比较显然的最小割，即删去最少的边，使得$x$和$y$通过小于和大于给定边边权的边都不连通。
• 时间复杂度$O(Dinic(N,M))$。

【代码】

#include<bits/stdc++.h>

using namespace std;

#define MAXN  20005


#define MAXM  200005


#define INF   1e9

template <typename T> void read(T &x) {
  x = 0; int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x *= f;
}
struct edge {int dest, flow; unsigned home; };
int n, m, s, t;
int dist[MAXN], p[MAXN], q[MAXN];
unsigned curr[MAXN];
vector <edge> a[MAXN];
int x[MAXM], y[MAXM], z[MAXM];
bool bfs() {
  memset(dist, 0, sizeof(dist));
  static int q[MAXN];
  int l = 0, r = 0;
  dist[s] = 1, q[0] = s;
  while (l <= r) {
      int tmp = q[l++];
      for (unsigned i = 0; i < a[tmp].size(); i++)
          if (a[tmp][i].flow != 0 && dist[a[tmp][i].dest] == 0) {
              dist[a[tmp][i].dest] = dist[tmp] + 1;
              q[++r] = a[tmp][i].dest;
          }
  }
  return dist[t] != 0;
}
int dinic(int pos, int limit) {
  if (pos == t) return limit;
  int used = 0, tmp;
  for (unsigned &i = curr[pos]; i < a[pos].size(); i++)
      if (a[pos][i].flow != 0 && dist[pos] + 1 == dist[a[pos][i].dest] && (tmp = dinic(a[pos][i].dest, min(limit - used, a[pos][i].flow)))) {
          used += tmp;
          a[pos][i].flow -= tmp;
          a[a[pos][i].dest][a[pos][i].home].flow += tmp;
          if (used == limit) return limit;
      }
  return used;
}
void addedge(int s, int t, int flow) {
  a[s].push_back((edge) {t, flow, a[t].size()});
  a[t].push_back((edge) {s, flow, a[s].size() - 1});
}
int main() {
  read(n), read(m);
  for (int i = 1; i <= m; i++)
      read(x[i]), read(y[i]), read(z[i]);
  int tx, ty, tl, ans = 0;
  read(tx), read(ty), read(tl);
  s = tx, t = ty;
  for (int i = 1; i <= m; i++)
      if (z[i] < tl) addedge(x[i], y[i], 1);
  while (bfs()) {
      memset(curr, 0, sizeof(curr));
      ans += dinic(s, INF);
  }
  for (int i = 1; i <= n; i++)
      a[i].clear();
  for (int i = 1; i <= m; i++)
      if (z[i] > tl) addedge(x[i], y[i], 1);
  while (bfs()) {
      memset(curr, 0, sizeof(curr));
      ans += dinic(s, INF);
  }
  cout << ans << endl;
  return 0;
}