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
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
//#pragma GCC optimize(2)
//#include <bits/stdc++.h>
#include <iostream>
#include <algorithm>
#include <cstdio>
#define rush() int T;cin>>T;while(T--)
#define go(a) while(cin>>a)
#define ms(a,b) memset(a,b,sizeof a)
#define E 1e-8
using namespace std;
typedef long long ll;
inline bool read(ll &num)
{char in;bool IsN=false;
in=getchar();if(in==EOF) return false;while(in!='-'&&(in<'0'||in>'9')) in=getchar();if(in=='-'){ IsN=true;num=0;}else num=in-'0';while(in=getchar(),in>='0'&&in<='9'){num*=10,num+=in-'0';}if(IsN) num=-num;return true;}
const int N=300+5;
ll n,m,t,_;
int i,j,k;
ll a[N];
ll p[N][N];
bool flag;
int main()
{
for(;read(n);){
for(i=1;i<=n;i++){
for(j=1;j<=n;j++){
read(p[i][j]);
p[j][i]=p[i][j];
}
}
ll ans=0;
for(i=1;i<=n;i++){
for(j=1;j<=n;j++){
flag=1;
for(k=1;k<=n;k++){
if(i==j || j==k || k==i) continue;
if(p[i][j]>p[i][k]+p[k][j]){
puts("-1");
return 0;
}
else
if(p[i][j]==p[i][k]+p[k][j])
flag=0;
}
if(flag) ans+=p[i][j];//人们想去中间城市，所以去掉p[i][j]直通路
}
}
cout<<ans/2<<endl;//i，j都是从1---n遍历，谁都可以是终点，谁都可以是起点
}
return 0;
}