UVa 10987 - Antifloyd

Summary
Given length of shortest path between every pair of vertices in an undirected graph with positive edge weights, construct such a graph, having minimum number of edges.

Explanation
Let the vertices be numbered from 1 to $$n$$, and let $$d_{ij}$$ be length of shortest path between $$i$$ and $$j$$.

First thing we need to do, is to check whether the graph actually exists, i.e. are the given $$d_{ij}$$'s lengths of shortest paths? If there exists three (distinct) vertices $$i$$, $$j$$, $$k$$, such that $$d_{ik}+d_{kj}<d_{ij}$$, then obviously the graph doesn't exist, because by concatenating shortest paths from $$i$$ to $$k$$ and from $$k$$ to $$j$$ we get a path from $$i$$ to $$j$$, of length less than $$d_{ij}$$.

If no such vertices exist, then we can always construct a graph, an example is a complete graph $$K_n$$, with weight of edge $$(i, j)$$ equal to $$d_{ij}$$. However it has $$n(n-1)/2$$ edges, can we do better?

Consider some pair of vertices $$i$$ and $$j$$. Shortest path between them is either a single edge $$(i, j)$$ of weight $$d_{ij}$$, or a sequence of two or more edges each of weight strictly less than $$d_{ij}$$ (due to assumption, that weight of every edge is greater than zero.) In the latter case, there exists some vertex $$k$$, different from $$i$$ and $$j$$, such that $$d_{ik}+d_{kj}=d_{ij}$$. If no such $$k$$ exists, the only option is to include the edge $$(i, j)$$ to our graph. Otherwise we can always omit edge $$(i, j)$$ -- there has to be a path from $$i$$ to $$j$$ in the final graph through a sequence of edges of length less than $$d_{ij}$$ (this can be formally proven by induction on $$d_{ij}$$.)

An algorithm solving the problem follows (in pseudo-code). Input: n = number of vertices, d[i][j] = shortest path lengths' matrix.

for i = 1 to n	for j = 1 to n		for k = 1 to n			if d[i][k]+d[k][j] < d[i][j]: return "Need better measurements."

for i = 1 to n	for j = 1 to n		flag = true for k = 1 to n			if k != i and k != j and d[i][k]+d[k][j]==d[i][j] flag = false if flag = true: add edge (i, j) of weight d[i][j] to the graph

Input
2 3 100 200 100 3 100 300 100

Output
Case #1: 2 1 2 100 2 3 100

Case #2: Need better measurements.