Graph Theory classic examples Awards

BZOJ 1614 to set up a telephone line

The maximum value of the minimum! ? Bipartite

Dichotomous answer:

Suppose takes mid-membered, can be used to see if the cost of the mid-membered 1 ~ n.

For w <mid, without free, 0

For w> mid must be free, 1

From there Can 1 ~ n <k Right side bar

With more than 0/1 of the principles established chart.

For the most short-circuit (short-circuit is most a few will contain a few, and containing a minimum of 1), if disN <= k is feasible

BZOJ1083 busy city:

https://www.luogu.org/problemnew/show/P2330

Minimum spanning tree run again on the line;

Violence O (C _M k) will obviously be fried

How to solve the problem of k = 1

Lift it out:

Copy the original again, and then even have the right to Collage 0, run the shortest path;

K edges:

Copy over k, 1 from layer 0 to layer k-w shortest! ?

Finally into 1 => (k-1) * n + n is shortest;

int d（int k，int i）{

return (k-1)*n+i

}

Wormhole: negative-side, the path: the positive-side;

Negative ring => You can return to the past;

Forfeit ring: if a click-throughs team> = n; has a negative ring; you can return to the past

On average, average into two teams on it; the number of times each point into the team is relatively small;

Suppose a small, but may be card number into the team assumed> = a constant, it is possible to have negative ring;

Single-source shortest run twice, respectively A, B as a starting point, intermediate point to find a minimum of x, such that the smallest dis (A, x) + dis (B, x), wherein:

Built map:

For each set, a new node, all the nodes in the collection to the new node is connected to one side of the right side of t (t is an arbitrary two points within the set of mutually time), by the new node is connected to a set point to the right side edge 0; then ↑

Intuitive feel dfs according sequential operation, must be optimal;

Dfs sequence according to a sequence of rows of treasure, the LCA is obtained between each two adjacent, then find the shortest path, and finally requires first and last of the shortest path;

Insert: insert in a, b a c satisfy a <c <b;

- the shortest path (a, b) + the shortest path (a, c) + the shortest path (b, c);