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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdint.h> // uint64_t

// 300*11
#define BUF_SIZE 3500
#define COST_MAX 0x7f7f7f7f

int get_int(void) {
  int num;
#ifdef BUF_SIZE
  char line[BUF_SIZE];
  if(!fgets(line, BUF_SIZE, stdin)) return 0;
  sscanf(line, "%d", &num);
#else
#error
#endif
  return num;
}

// <arr[0]> <arr[1]> .. <arr[size-1]>
int fget_array(int *arr, int size) {
#ifdef BUF_SIZE
    char line[BUF_SIZE];
    char *tmpbuf = line;
    int i;
    if(!fgets(line, BUF_SIZE, stdin)) return -1;
    for(i = 0; i < size; i++) {
        char *tmp = strtok(tmpbuf, " ");
        arr[i] = strtol(tmp, NULL, 10);
        tmpbuf = NULL;
    }
#else
#error
#endif
    return 0;
}

#define VERTEX_MAX 300

int main(void) {
    int cities;
    int graph[VERTEX_MAX][VERTEX_MAX];
    int roads[VERTEX_MAX][VERTEX_MAX];
    memset(roads, 0x7f, sizeof(int)*VERTEX_MAX*VERTEX_MAX);
    cities = get_int();

    int i, j, k;
    for(i = 0; i < cities; i++) {
        fget_array(graph[i], cities);
    }

    // Warshall Floyd (connect step)
    for(k = 0; k < cities; k++) {
        for(i = 0; i < cities; i++) {
            if(i == k) continue;
            for(j = 0; j < cities; j++) {
                if(i == j || j == k) continue;
                // < 2-billion
                int new = graph[i][k] + graph[k][j];
                // ensure that all roads are shortest path
                if(new < graph[i][j]) {
                    printf("-1\n");
                    return 0;
                }
                // building 2 roads are more convenient
                if(new == graph[i][j]) {
                    roads[i][k] = graph[i][k];
                    roads[k][j] = graph[k][j];
                }
            }
        }
    }

    // Warshall Floyd (remove step)
    for(k = 0; k < cities; k++) {
        for(i = 0; i < cities; i++) {
            if(i == k) continue;
            for(j = 0; j < cities; j++) {
                // ensure that i, j and k are all different
                if(i == j || j == k) continue;
                // < 2-billion
                int new = graph[i][k] + graph[k][j];
                // i<->j can be removed because of more convenient paths between i and j.
                if(new == graph[i][j]) {
                    roads[i][j] = 0;
                }
            }
        }
    }

    // we can't help building road
    for(i = 0; i < cities; i++) {
        for(j = 0; j < cities; j++) {
            if(roads[i][j] == COST_MAX) {
                roads[i][j] = graph[i][j];
            }
        }
    }

    uint64_t ans = 0;
    for(i = 0; i < cities; i++) {
        for(j = i+1; j < cities; j++) {
#ifdef DEBUG
            printf("%d, %d -> %d\n", i,j,roads[i][j]);
#endif
            ans += (uint64_t)roads[i][j];
        }
    }
    printf("%llu\n", ans);
    return 0;
}