UVa 10724 - Road Construction

Summary
Given is an undirected connected graph, which represents a city. Graph's vertices are points on 2D plane, numbered from 1 to $$n$$. Weight of an edge (when it exists) is equal to the distance between points.

Find a pair of vertices $$(u,v)$$, such that:
 * $$u < v$$.
 * The edge $$(u,v)$$ isn't present in the graph.
 * $$C_{u,v} = \sum_{i=1}^n \sum_{j=i+1}^n (d_{i,j} - d'_{i,j})$$ is maximized. $$d_{i,j}$$ is the shortest distance between vertices $$i,j$$ in the original graph. $$d'_{i,j}$$ is the shortest distance between them, when the edge $$(u,v)$$ is added to the graph.
 * $$C_{u,v} > 1$$
 * When there are several pairs, satisfying above conditions, choose the one with smallest straight-line distance between vertices.
 * If there's still a tie, favor a pair with smaller $$u$$, and smaller $$v$$.

Explanation
A simple $$O(n^4)$$ algorithm is enough to get accepted. First, use Floyd-Warshall to compute the shortest distances in the original graph, and then check all pairs $$(u,v)$$. For each pair compute $$C_{u,v}$$ from its definition.

The shortest path between vertices $$i$$ and $$j$$ after the edge $$(u,v)$$ is added is equal to:

$$min(d_{i,j},\ d_{i,u}+w_{u,v}+d_{v,j},\ d_{i,v}+w_{v,u}+d_{u,j})$$

where $$w_{u,v}$$ is the straight-line distance between vertices $$u$$ and $$v$$.

Input
4 6 0 0 0 2 2 0 2 2 1 2 1 3 1 4 2 3 2 4 3 4 4 4 0 0 0 2 2 0 2 2 1 2 2 3 3 4 4 1 4 3 0 0 0 2 2 0 2 2 1 2 2 3 3 4 0 0

Output
No road required 1 3 1 3