【BZOJ2229】【ZJOI2011】最小割

【题目链接】

• 点击打开链接

【思路要点】

• 建出GomoryHu-Tree ，剩余部分用DFS即可解决。
• 时间复杂度$O(T*(N*Dinic(N,M)+NQ))$。

【代码】

#include<bits/stdc++.h>

using namespace std;

#define MAXN  205


#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; };
struct treeedge {int dest, index; };
int n, m, s, t, Q, cnt, value;
int dist[MAXN], p[MAXN], q[MAXN];
unsigned curr[MAXN];
bool visited[MAXN];
vector <edge> a[MAXN];
vector <treeedge> b[MAXN];
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});
}
void ClearGraph() {
  for (int i = 1; i <= n; i++)
  for (unsigned j = 0; j < a[i].size(); j++)
      a[i][j].flow = a[a[i][j].dest][a[i][j].home].flow = (a[i][j].flow + a[a[i][j].dest][a[i][j].home].flow ) / 2;
}
void solve(int l, int r) {
  if (l == r) return;
  s = p[l], t = p[r];
  int flow = 0;
  ClearGraph();
  while (bfs()) {
      memset(curr, 0, sizeof(curr));
      flow += dinic(s, INF);
  }
  b[s].push_back((treeedge) {t, flow});
  b[t].push_back((treeedge) {s, flow});
  int tl = l - 1, tr = r + 1;
  for (int i = l; i <= r; i++)
      if (dist[p[i]]) q[++tl] = p[i];
      else q[--tr] = p[i];
  for (int i = l; i <= r; i++)
      p[i] = q[i];
  solve(l, tl); solve(tr, r);
}
void work(int pos) {
  visited[pos] = true; cnt++;
  for (unsigned i = 0; i < b[pos].size(); i++)
      if (!visited[b[pos][i].dest] && b[pos][i].index > value) work(b[pos][i].dest);
}
int main() {
  int T; read(T);
  while (T--) {
      read(n), read(m);
      for (int i = 1; i <= n; i++) {
          a[i].clear();
          b[i].clear();
      }
      for (int i = 1; i <= m; i++) {
          int x, y, z;
          read(x), read(y), read(z);
          if (x != y) addedge(x, y, z);
      }
      for (int i = 1; i <= n; i++)
          p[i] = i;
      solve(1, n);
      read(Q);
      for (int i = 1; i <= Q; i++) {
          memset(visited, false, sizeof(visited));
          read(value);
          int ans = n * (n - 1) / 2;
          for (int j = 1; j <= n; j++) {
              if (visited[j]) continue;
              cnt = 0; work(j);
              ans -= cnt * (cnt - 1) / 2;
          }
          printf("%d\n", ans);
      }
      printf("\n");
  }
  return 0;
}