问题 C: Restoring Road Network
时间限制: 1 Sec 内存限制: 128 MB
提交: 790 解决: 161
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题目描述
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
输入
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
输出
If there exists no network that satisfies the condition, print -1. If it exists, print the shortest possible total length of the roads.
样例输入
3 0 1 3 1 0 2 3 2 0
样例输出
3
提示
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.
题目大意：有一个 N*N 的矩阵，每一行 i 和 每一列 j 对应的坐标表示从 i 点到 j 点的路径距离，路径是双向的。判断所给路径是否为从 i 到 j 的最短路径，如果不是输出 -1；如果是则求出所需路径的最小距离。
解题思路：因为 N 数值较小，因此我们先用 Floyd 算法求出每两点中转路径之间的最短距离（不包含直接从 i 到 j 的路径），然后遍历所有路径。如果存在中转路径比直接路径数值小，则输出 -1；如果中转路径比直接路径大，我们加上直接路径的距离；如果相等，因为题目要求最小距离，所以不操作，直接进入下一个循环。
参考代码：
#include<bits/stdc++.h>
using namespace std;
long long a[310][310];
int main()
{
int n;
while(cin>>n)
{
int flg=0;
for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++)
cin>>a[i][j];
long long ans=0;
for(int i=1; i<=n; i++)
for(int j=i+1; j<=n; j++)
{
long long ans1=0x3f3f3f3f;
for(int k=1; k<=n; k++){
if(k==i||k==j) continue;
ans1=min(a[i][k]+a[k][j],ans1);
}
if(ans1==a[i][j]) continue;
if(ans1>a[i][j]) ans+=a[i][j];
if(ans1<a[i][j]) flg=1;
}
if(flg) cout<<-1<<endl;
else
cout<<ans<<endl;
}
}