Restoring Road Network

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

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

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

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

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define LL long long
 4 #define eps 1e-8
 5 #define MX 305
 6 
 7 int n;
 8 int G[MX][MX];
 9 
10 int main()
11 {
12     while (scanf("%d",&n)!=EOF)
13     {
14         for (int i=1;i<=n;i++)
15             for (int j=1;j<=n;j++)
16                 scanf("%d",&G[i][j]);
17         LL ans =0;
18         bool ok=1;
19         for (int i=1;i<=n;i++)
20         {
21             for (int j=1;j<=n;j++)
22             {
23                 if (i==j) continue;
24                 bool need=1;
25                 for (int k=1;k<=n;k++)
26                 {
27                     if (k==i||k==j) continue;
28                     if (G[i][j]>G[i][k]+G[k][j])  ok=0;
29                     if (G[i][j]==G[i][k]+G[k][j]) need=0;
30                 }
31                 if (need) ans+=G[i][j];
32             }
33         }
34         if (ok) printf("%lld\n",ans/2);
35         else printf("-1\n");
36     }
37     return 0;
38 }

View Code