Problem:
Score : 500 points

Problem Statement

In Takahashi Kingdom, which once existed, there are N cities, and some pairs of cities are connected bidirectionally by roads.
The following are known about the road network:

People traveled between cities only through roads. It was possible to reach any city from any other city, via intermediate cities if necessary.

Different roads may have had different lengths, but all the lengths were positive integers.

Snuke the archeologist found a table with N rows and N columns, A, in the ruin of Takahashi Kingdom.
He thought that it represented the shortest distances between the cities along the roads in the kingdom.

Determine whether there exists a road network such that for each u and v, the integer A_{u, v} at the u-th row and v-th column of A is equal to the length of the shortest path from City u to City v.
If such a network exist, find the shortest possible total length of the roads.

Constraints

1 \leq N \leq 300

If i ≠ j, 1 \leq A_{i, j} = A_{j, i} \leq 10^9.

A_{i, i} = 0

Inputs

Input is given from Standard Input in the following format:

N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}
...
A_{N, 1} A_{N, 2} ... A_{N, N}

Outputs

If there exists no network that satisfies the condition, print -1.
If it exists, print the shortest possible total length of the roads.

Sample Input 1

3
0 1 3
1 0 2
3 2 0

Sample Output 1

3

The network below satisfies the condition:

City 1 and City 2 is connected by a road of length 1.

City 2 and City 3 is connected by a road of length 2.

City 3 and City 1 is not connected by a road.

Sample Input 2

3
0 1 3
1 0 1
3 1 0

Sample Output 2

-1

As there is a path of length 1 from City 1 to City 2 and City 2 to City 3, there is a path of length 2 from City 1 to City 3.
However, according to the table, the shortest distance between City 1 and City 3 must be 3.

Thus, we conclude that there exists no network that satisfies the condition.

Sample Input 3

5
0 21 18 11 28
21 0 13 10 26
18 13 0 23 13
11 10 23 0 17
28 26 13 17 0

Sample Output 3

82

Sample Input 4

3
0 1000000000 1000000000
1000000000 0 1000000000
1000000000 1000000000 0

Sample Output 4

3000000000

Solution in LuaJIT (2.0.4) (judge verdict: Time Limit Exceeded)

local function uf_initialize(n)
  local parent = {}
  for i = 1, n do parent[i] = i end
  return parent
end

local function uf_findroot(idx, parent)
  local idx_update = idx
  while parent[idx] ~= idx do
    idx = parent[idx]
  end
  while parent[idx_update] ~= idx do
    parent[idx_update], idx_update = idx, parent[idx_update]
  end
  return idx
end

local n = io.read("*n")
local edge = {}
for i = 1, n do
  for j = 1, n do
    local c = io.read("*n")
    if i < j then
      local tmp = {}
      tmp.c, tmp.i1, tmp.i2 = c, i, j
      table.insert(edge, tmp)
    end
  end
end

local e = {}
for i = 1, n do e[i] = {} end

local function getlen(start_idx, dst_idx)
  local taskstate = {}
  for i = 1, n do taskstate[i] = false end
  local tasks = {}
  local tasknum = 0
  local done = 0

  local len = {}
  for i = 1, n do len[i] = 300000000000 end
  len[start_idx] = 0

  local function addtask(idx)
    if not taskstate[idx] then
      taskstate[idx] = true
      tasknum = tasknum + 1
      tasks[tasknum] = idx
    end
  end
  addtask(start_idx)

  while done < tasknum do
    done = done + 1
    local src = tasks[done]
    taskstate[src] = false
    for dst, cost in pairs(e[src]) do
      if len[src] + cost < len[dst] then
        len[dst] = len[src] + cost
        addtask(dst)
      end
    end
  end
  return len[dst_idx]
end

table.sort(edge, function(a, b) return a.c < b.c end)
local parent = uf_initialize(n)
local totalcost = 0
local flg = true
local linenum = #edge
for i = 1, linenum do
  local c, i1, i2 = edge[i].c, edge[i].i1, edge[i].i2
  local r1, r2 = uf_findroot(i1, parent), uf_findroot(i2, parent)
  parent[i1], parent[i2] = r1, r2
  if r1 ~= r2 then
    parent[r2] = r1
    parent[i2] = r1
    totalcost = totalcost + c
  else
    local len = getlen(i1, i2)
    if c < len then
      totalcost = totalcost + c
    elseif len < c then
      flg = false
    end
  end
  e[i1][i2], e[i2][i1] = c, c
end
print(flg and totalcost or -1)