Restoring Road Network

D - Restoring Road Network

---

Time limit : 2sec / Memory limit : 256MB

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 Au,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≤N≤300
• If i≠j, 1≤Ai,j=Aj,i≤109.
• Ai,i=0

---

Inputs

Input is given from Standard Input in the following format:

N
A1,1 A1,2 … A1,N
A2,1 A2,2 … A2,N
…
AN,1 AN,2 … AN,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

Copy

3
0 1 3
1 0 2
3 2 0

Sample Output 1

Copy

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

Copy

3
0 1 3
1 0 1
3 1 0

Sample Output 2

Copy

-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

Copy

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

Copy

82

---

Sample Input 4

Copy

3
0 1000000000 1000000000
1000000000 0 1000000000
1000000000 1000000000 0

Sample Output 4

Copy

3000000000

//题意:给出一个 n * n 的最短路表，问此表需要最少连通多少边多少才能实现。、

//显然，对于每对点都要考虑，如果，可以通过第三方点实现，就用第三方，否则，只能连本身的边

 #include<bits/stdc++.h>
 using namespace std;
 #define LL long long
 #define eps 1e-8
 #define MX 305

 int n;
 int G[MX][MX];

 int main()
 {
     while (scanf("%d",&n)!=EOF)
     {
         for (int i=;i<=n;i++)
             for (int j=;j<=n;j++)
                 scanf("%d",&G[i][j]);
         LL ans =;
         bool ok=;
         for (int i=;i<=n;i++)
         {
             for (int j=;j<=n;j++)
             {
                 if (i==j) continue;
                 bool need=;
                 for (int k=;k<=n;k++)
                 {
                     if (k==i||k==j) continue;
                     if (G[i][j]>G[i][k]+G[k][j])  ok=;
                     if (G[i][j]==G[i][k]+G[k][j]) need=;
                 }
                 if (need) ans+=G[i][j];
             }
         }
         if (ok) printf("%lld\n",ans/);
         else printf("-1\n");
     }
     return ;
 }