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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

#define li long long int
#define rep(i, to) for (li i = 0; i < ((li)(to)); i++)
#define repp(i, start, to) for (li i = (li)(start); i < ((li)(to)); i++)
#define pb push_back
#define sz(v) ((li)(v).size())
#define bgn(v) ((v).begin())
#define eend(v) ((v).end())
#define allof(v) (v).begin(), (v).end()
#define dodp(v, n) memset(v, (li)n, sizeof(v))
#define bit(n) (1ll << (li)(n))
#define mp(a, b) make_pair(a, b)
#define rin rep(i, n)
#define EPS 1e-12
#define ETOL 1e-8
#define MOD 1000000007
typedef pair<li, li> PI;

#define INF bit(60)

#define DBGP 1

#define idp if (DBGP)
#define F first
#define S second
#define p2(a, b) idp cout << a << "\t" << b << endl
#define p3(a, b, c) idp cout << a << "\t" << b << "\t" << c << endl
#define p4(a, b, c, d) idp cout << a << "\t" << b << "\t" << c << "\t" << d << endl
#define p5(a, b, c, d, e) idp cout << a << "\t" << b << "\t" << c << "\t" << d << "\t" << e << endl
#define p6(a, b, c, d, e, f) idp cout << a << "\t" << b << "\t" << c << "\t" << d << "\t" << e << "\t" << f << endl
#define p7(a, b, c, d, e, f, g) idp cout << a << "\t" << b << "\t" << c << "\t" << d << "\t" << e << "\t" << f << "\t" << g << endl
#define p8(a, b, c, d, e, f, g, h) idp cout << a << "\t" << b << "\t" << c << "\t" << d << "\t" << e << "\t" << f << "\t" << g << "\t" << h << endl
#define p9(a, b, c, d, e, f, g, h, i) idp cout << a << "\t" << b << "\t" << c << "\t" << d << "\t" << e << "\t" << f << "\t" << g << "\t" << h << "\t" << i << endl
#define p10(a, b, c, d, e, f, g, h, i, j) idp cout << a << "\t" << b << "\t" << c << "\t" << d << "\t" << e << "\t" << f << "\t" << g << "\t" << h << "\t" << i << "\t" << j << endl
#define foreach(it, v) for (__typeof((v).begin()) it = (v).begin(); it != (v).end(); ++it)
#define p2p(x) idp p2((x).F, (x).S)
#define dump(x, n)                         \
    idp                                    \
    {                                      \
        rep(i, n) { cout << x[i] << " "; } \
        puts("");                          \
    }
#define dump2(x, n)                                                     \
    idp                                                                 \
    {                                                                   \
        rep(i, n) { cout << "[" << x[i].F << " , " << x[i].S << "] "; } \
        puts("");                                                       \
    }
#define dumpi(x)                  \
    idp                           \
    {                             \
        foreach (it, x)           \
        {                         \
            cout << (*it) << " "; \
        }                         \
        puts("");                 \
    }
#define dumpi2(x)                                               \
    idp                                                         \
    {                                                           \
        foreach (it, x)                                         \
        {                                                       \
            cout << "[" << (it)->F << " , " << (it)->S << "] "; \
        }                                                       \
        puts("");                                               \
    }

#define read2d(a, w, h) rep(i, h) rep(j, w) cin >> a[i][j]
#define dump2d(a, w, h)                   \
    rep(i, h)                             \
    {                                     \
        rep(j, w) cout << a[i][j] << " "; \
        puts("");                         \
    }

typedef pair<li, li> PI;

li table[333][333];
li a[333][333];

// {distance, {from, to}}
vector<pair<li, PI>> distances;
li n;

inline void update(li d, li from, li to) {
    rep(i, n) {
        table[i][to] = min(table[i][to], table[i][from] + d);
        table[i][from] = min(table[i][from], table[i][to] + d);
        table[to][i] = table[i][to];
        table[from][i] = table[i][from];
    }
}

int main(void)
{
    cin >> n;
    rin{
        rep(j, n) {
            cin >> a[i][j];
            table[i][j] = INF;
            if (i == j) {
                table[i][j] = 0;
            } else {
                distances.pb({a[i][j], {i, j}});
            }
        }
    }

    sort(allof(distances));
    li res = 0;

    for (auto && item : distances) {
        li d = item.F;
        li from = item.S.F;
        li to = item.S.S;
        if (table[from][to] < d) {
            puts("-1");
            return 0;
        }
        if (table[from][to] > d) {
            table[from][to] = d;
            table[to][from] = d;
            update(d, from, to);
            //p3(d, from, to);
            //dump2d(table, n, n);
            //puts("");
            res += d;
        }
    }


    rin{
        rep(j, n) {
            if (a[i][j] != table[i][j]) {
                puts("-1");
                return 0;
            }
        }
    }
    cout << res << endl;

    return 0;
}