Hexo
0%

最小树形图

Posted on In ,
Symbols count in article: 3.7k Reading time ≈ 3 mins.

最小树形图

有向图上的最小生成树称为最小树形图

Edmonds算法

O(nm)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
bool solve() {
  ans = 0;
  int u, v, root = 0;
  for (;;) {
    f(i, 0, n) in[i] = 1e100;
    f(i, 0, m) {
      u = e[i].s;
      v = e[i].t;
      if (u != v && e[i].w < in[v]) {
        in[v] = e[i].w;
        pre[v] = u;
      }
    }
    f(i, 0, m) if (i != root && in[i] > 1e50) return 0;
    int tn = 0;
    memset(id, -1, sizeof id);
    memset(vis, -1, sizeof vis);
    in[root] = 0;
    f(i, 0, n) {
      ans += in[i];
      v = i;
      while (vis[v] != i && id[v] == -1 && v != root) {
        vis[v] = i;
        v = pre[v];
      }
      if (v != root && id[v] == -1) {
        for (int u = pre[v]; u != v; u = pre[u]) id[u] = tn;
        id[v] = tn++;
      }
    }
    if (tn == 0) break;
    f(i, 0, n) if (id[i] == -1) id[i] = tn++;
    f(i, 0, m) {
      u = e[i].s;
      v = e[i].t;
      e[i].s = id[u];
      e[i].t = id[v];
      if (e[i].s != e[i].t) e[i].w -= in[v];
    }
    n = tn;
    root = id[root];
  }
  return ans;
}

Tarjan算法

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
#define maxn 102
#define INF 0x3f3f3f3f

struct UnionFind {
  int fa[maxn << 1];

  UnionFind() { memset(fa, 0, sizeof(fa)); }

  void clear(int n) { memset(fa + 1, 0, sizeof(int) * n); }

  int find(int x) { return fa[x] ? fa[x] = find(fa[x]) : x; }

  int operator[](int x) { return find(x); }
};

struct Edge {
  int u, v, w, w0;
};

struct Heap {
  Edge *e;
  int rk, constant;
  Heap *lch, *rch;

  Heap(Edge *_e) : e(_e), rk(1), constant(0), lch(NULL), rch(NULL) {}

  void push() {
    if (lch) lch->constant += constant;
    if (rch) rch->constant += constant;
    e->w += constant;
    constant = 0;
  }
};

Heap *merge(Heap *x, Heap *y) {
  if (!x) return y;
  if (!y) return x;
  if (x->e->w + x->constant > y->e->w + y->constant) swap(x, y);
  x->push();
  x->rch = merge(x->rch, y);
  if (!x->lch || x->lch->rk < x->rch->rk) swap(x->lch, x->rch);
  if (x->rch)
    x->rk = x->rch->rk + 1;
  else
    x->rk = 1;
  return x;
}

Edge *extract(Heap *&x) {
  Edge *r = x->e;
  x->push();
  x = merge(x->lch, x->rch);
  return r;
}

vector<Edge> in[maxn];
int n, m, fa[maxn << 1], nxt[maxn << 1];
Edge *ed[maxn << 1];
Heap *Q[maxn << 1];
UnionFind id;

void contract() {
  bool mark[maxn << 1];
  // 将图上的每一个结点与其相连的那些结点进行记录。
  for (int i = 1; i <= n; i++) {
    queue<Heap *> q;
    for (int j = 0; j < in[i].size(); j++) q.push(new Heap(&in[i][j]));
    while (q.size() > 1) {
      Heap *u = q.front();
      q.pop();
      Heap *v = q.front();
      q.pop();
      q.push(merge(u, v));
    }
    Q[i] = q.front();
  }
  mark[1] = true;
  for (int a = 1, b = 1, p; Q[a]; b = a, mark[b] = true) {
    // 寻找最小入边以及其端点，保证无环。
    do {
      ed[a] = extract(Q[a]);
      a = id[ed[a]->u];
    } while (a == b && Q[a]);
    if (a == b) break;
    if (!mark[a]) continue;
    // 对发现的环进行收缩，以及环内的结点重新编号，总权值更新。
    for (a = b, n++; a != n; a = p) {
      id.fa[a] = fa[a] = n;
      if (Q[a]) Q[a]->constant -= ed[a]->w;
      Q[n] = merge(Q[n], Q[a]);
      p = id[ed[a]->u];
      nxt[p == n ? b : p] = a;
    }
  }
}

ll expand(int x, int r);

ll expand_iter(int x) {
  ll r = 0;
  for (int u = nxt[x]; u != x; u = nxt[u]) {
    if (ed[u]->w0 >= INF)
      return INF;
    else
      r += expand(ed[u]->v, u) + ed[u]->w0;
  }
  return r;
}

ll expand(int x, int t) {
  ll r = 0;
  for (; x != t; x = fa[x]) {
    r += expand_iter(x);
    if (r >= INF) return INF;
  }
  return r;
}

void link(int u, int v, int w) { in[v].push_back({u, v, w, w}); }

int main() {
  int rt;
  scanf("%d %d %d", &n, &m, &rt);
  for (int i = 0; i < m; i++) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    link(u, v, w);
  }
  // 保证强连通
  for (int i = 1; i <= n; i++) link(i > 1 ? i - 1 : n, i, INF);
  contract();
  ll ans = expand(rt, n);
  if (ans >= INF)
    puts("-1");
  else
    printf("%lld\n", ans);
  return 0;
}