bzoj-3118 Orz the MST

Test instructions

An undirected connected graph is given, and one of the spanning trees is specified;

Each edge has a weight of Vali, if the increase of the weight of 1 to spend the AI, reduce the cost of 1 bi;

Now to make the specified spanning tree a minimum spanning tree, the minimum cost is calculated;

n<=300,m<=1000;

Exercises

A linear programming comparison of God's topic, brush in front of the comparison of the water will not brush;

First there is a very obvious thing (I did not see it), the edge of the tree must reduce the weight, not the edge of the tree must increase the weight value;

Then consider for a spanning tree to be the minimum to meet the conditions, that is, the constraint conditions of the subject;

Like the Tarjan algorithm, each non-tree edge generates a ring on the tree;

And if some edge of the ring is larger than it, then the ring can be disconnected from that edge, and the spanning tree is smaller;

In other words, for a non-tree edge, the top of the ring is less than equal to it;

Find the process of constraint I write seems rather stupid, deep search to remember a bunch of re-LCA;

But in any case the worst but n*m will not be so tle;

Set XI as the first edge of the change amount, then the edge of the tree is the amount of reduction, non-tree edge for the increase of a large number;

Available equation: Vali+xi<=valj-xj (i is a tree edge J is not a tree edge, and i,j on a ring);

Move Item:xi+xj<=valj-vali;

The current linear programming is:

Min∑costi*xi

Xi+xj<=valj-vali

Obviously it does not exist the basic feasible solution, then does some deformation;

First, the target function is reversed:

Max∑-costi*xi

Xi+xj<=valj-vali

Then the dual principle!

Max∑ (Valj-vali) *x (j,i)

%*&^%*>=-costi

And then the inequality variable number:

Max∑ (Valj-vali) *x (j,i)

-%*&^%*<=costi

Now it can be found that the linear programming is already the standard type;

And the cost in the title is positive, that is, the linear programming has a basic feasible solution;

Then on the simple type directly engage in this thing, the nature of the topic also guarantees that this thing will not be unbounded;

Time complexity O (Simple Type (n*m,m));

Obviously it will be T-dead, but the constraints will not have the worst case n*m so much.

So it can still be n*m, the array of the open to 4000 can A;

Code:

#include <vector> #include <math.h> #include <stdio.h> #include <string.h> #include < Algorithm> #define N 310 #define M 1100 using namespace std; Const double EPS = 1e-8; Const double INF = 1e100;     struct Node {int n, m;     Double a[m][m << 2], b[m], c[m << 2], V; int find () {for (int i = 1; I <= n; i++) {if (C[i] > EPS) return i         ;     } return 0;         } void Rotate (int l, int e) {b[l]/= a[l][e];         for (int i = 1; I <= n; i++) {if (i! = e) a[l][i]/= a[l][e];         } A[l][e] = 1/a[l][e];             for (int i = 1; I <= m; i++) {if (i = = L | | fabs (a[i][e]) < EPS) continue;             B[i]-= b[l] * A[i][e]; for (int j = 1; J <= N; j + +) {if (j! = e) a[i][j]-= a[l][j] * A[i][e]             ; } A[i][e] *=-a[l][e];         } v + = c[e] * B[l];         for (int i = 1; I <= n; i++) {if (i! = e) c[i]-= c[e] * A[l][i];     } C[e] *=-a[l][e];         } void Simplex () {int I, j, K, L, E;             while (E = find ()) {double Lim = INF;                 for (i = 1; I <= m; i++) {if (A[i][e] < EPS) continue;             if (Lim>b[i]/a[i][e]) Lim = B[i]/a[i][e], L = i;         } Rotate (L, E); }}}t; Vector<int>to[n], Val[n], cost[n], no[n]; vector<bool>cov[n]; int fa[n], prec[n], prev[n], preno[n], tim[n], deep[n], tot;     void Build (int x, int y, int val, int no) {if (Deep[x] < deep[y]) swap (x, y);         while (Deep[x]>deep[y]) {T.C[++T.N] =-(val-prev[x]);         T.A[NO][T.N] = 1;         T.A[PRENO[X]][T.N] = 1;     x = Fa[x];         } while (x! = y) {T.C[++T.N] =-(val-prev[x]);         T.A[NO][T.N] = 1;         T.A[PRENO[X]][T.N] = 1;         x = Fa[x];         T.C[++T.N] =-(val-prev[y]);         T.A[NO][T.N] = 1;         T.A[PRENO[Y]][T.N] = 1;     y = fa[y];     }} void Dfs (int x, int pre, int v, int num, int d) {int i, y;     Fa[x]=pre,prev[x] = V, preno[x] = num, tim[x] = ++tot, deep[x] = D; for (i = 0; i < to[x].size (); i++) {if (cov[x][i]&& (Y=to[x][i])!=pre) Dfs (y, x, val[x)     [i], no[x][i], D + 1); } for (i = 0; i < to[x].size (); i++) {if (!cov[x][i] && tim[y = to[x][i]] && Tim[y] &l T         Tim[x]) {Build (x, Y, Val[x][i], no[x][i]);     }}} int main () {int n, m, I, J, K, X, Y, V, F, a, B;     scanf ("%d%d", &n, &m);         for (i = 1; I <= m; i++) {scanf ("%d%d%d%d%d%d", &x, &y, &v, &f, &a, &b); To[x].push_back (y), Val[x].push_back (v), Cov[x].push_back (f), No[x].push_back (i);         To[y].push_back (x), Val[y].push_back (v), Cov[y].push_back (f), No[y].push_back (i);         if (f) t.b[i] = b;     else t.b[i] = A;     } DFS (1, 0, 0, 0, 1);     T.m=m;     T.simplex ();     printf ("%.0LF", T.V); return 0; }

bzoj-3118 Orz the MST