hdu 4034 Graph 已知最短路，求图中最少边数_已知最短路,求可能的图-优快云博客

本文链接：https://blog.youkuaiyun.com/kongming_acm/article/details/6771899

本文探讨了如何从给定的最短路径矩阵推导出原始图的最小边数，通过实例展示了求解过程，并指出在特定情况下无法构建相应的原始图。

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Problem Description

Everyone knows how to calculate the shortest path in a directed graph. In fact, the opposite problem is also easy. Given the length of shortest path between each pair of vertexes, can you find the original graph?

Input

The first line is the test case number T (T ≤ 100).
First line of each case is an integer N (1 ≤ N ≤ 100), the number of vertexes.
Following N lines each contains N integers. All these integers are less than 1000000.
The jth integer of ith line is the shortest path from vertex i to j.
The ith element of ith line is always 0. Other elements are all positive.

Output

For each case, you should output “Case k: ” first, where k indicates the case number and counts from one. Then one integer, the minimum possible edge number in original graph. Output “impossible” if such graph doesn't exist.

Sample Input

Sample Output

  
  
   
   Case 1: 6
Case 2: 4
Case 3: impossible

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=110;
int a[maxn][maxn];
int vis[maxn][maxn];
int n;
int main()
{
    int ci,pl=1;scanf("%d",&ci);
    while(ci--)
    {
        scanf("%d",&n);
        int ans=n*(n-1);
        memset(vis,0,sizeof(vis));
        int flag=1;
        for(int i=0;i<n;i++) for(int j=0;j<n;j++) scanf("%d",&a[i][j]);
        for(int k=0;k<n&&flag;k++)
        {
            for(int i=0;i<n&&flag;i++)
            {
                for(int j=0;j<n&&flag;j++)
                {
                    if(i!=j&&j!=k&&i!=k&&!vis[i][k]&&!vis[k][j]&&!vis[i][j])
                    {
                        if(a[i][j]==a[i][k]+a[k][j])
                        {
                            ans--;
                            vis[i][j]=1;
                        }
                        else if(a[i][j]>a[i][k]+a[k][j])
                        {
                            flag=0;break;
                        }
                    }
                }
            }
        }
        if(flag) printf("Case %d: %d\n",pl++,ans);
        else printf("Case %d: impossible\n",pl++);
    }
    return 0;
}