Graph_hdu_4034 (delete edge). java

GraphTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others) Total Submission (s): 1587 Accepted Submission (s ): 790 Problem DescriptionEveryone 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? InputThe 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. outputFor 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 Input330 1 11 0 11 1 030 1 3 4 0 27 3 030 1 41 0 24 2 0 Sample OutputCase 1: 6 Case 2: 4 Case 3: impossible SourceThe 36th ACM/ICPC Asia Regional Chengdu Site -- Online Contest Recommendlcy | We have carefully selected several similar problems for you: 4039 4038 4036 4033 4037

import java.io.InputStreamReader;  import java.util.Scanner;    public class Main{      public static void main(String[] args) {          Scanner input=new Scanner(new InputStreamReader(System.in));          int N=input.nextInt();          int e=1;          while(N-->0){              int n=input.nextInt();              int a[][]=new int[n+1][n+1];              for(int i=1;i<=n;i++){                  for(int j=1;j<=n;j++){                      a[i][j]=input.nextInt();                  }              }              boolean ok=false;              for(int i=1;i<=n;i++){                  for(int j=1;j<=n;j++){                      if(i==j)                          continue;                      for(int k=1;k<=n;k++){                          if(j==k)                              continue;                          if(a[k][i]!=0&&a[i][j]!=0){                              if(a[k][i]+a[i][j]==a[k][j])                                  a[k][j]=0;                              else if(a[k][i]+a[i][j]<a[k][j]){                                  System.out.println("Case "+(e++)+": impossible");                                  ok=true;                                  break;                              }                          }                      }                      if(ok)                          break;                  }                  if(ok)                      break;              }              if(ok)                  continue;              int sum=0;              for(int i=1;i<=n;i++){                  for(int j=1;j<=n;j++){                      if(a[i][j]!=0)                          sum++;                  }              }              System.out.println("Case "+(e++)+": "+sum);          }      }  }