HDU4034-Graph

Graph

                                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
                                                                                                       Total Submission(s): 2335    Accepted Submission(s): 1186

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

  

   3
3
0 1 1
1 0 1
1 1 0
3
0 1 3 
4 0 2
7 3 0
3
0 1 4
1 0 2
4 2 0
  

 

Sample Output

  

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

 

Source

The 36th ACM/ICPC Asia Regional Chengdu Site —— Online Contest

题意是：给我们两点间的最短路径（双向），让我们求原来图中的最少有多少条边。能取消边的条件是当一个点到达另一个点的最短路径等于借助于第三个点的最短路径；例如：1-->2的最短路径为4 ，2-->3的最短路径为3而1-->3的最短路径为7所以1-->3的这条边就可以去掉；如果题目给的1-->3的最短路径大于7，这就不可能；因为1-->3借助于2的最短路径是7

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int map[110][110],visit[110][110],ans,n;

int Floyd()
{
    for (int k=0;k<n;k++)
    {
        for (int i=0;i<n;i++)
        {
            for (int j=0;j<n;j++)
            {
                if (i!=j&&j!=k&&i!=k)
                {
                    if(!visit[i][j]&&(map[i][j]==map[i][k]+map[k][j]))
                    {
                        ans--;
                        visit[i][j]=1;
                    }
                    else if(map[i][j]>map[i][k]+map[k][j])
                        return 0;
                }
            }
        }
    }
    return 1;
}

int main()
{
    int cas=1,t;
    scanf("%d",&t);
    while (t--)
    {
        memset(visit,0,sizeof(visit));
        scanf("%d", &n);
        for (int i=0;i<n;i++)
        {
            for (int j=0;j<n;j++)
                scanf("%d",&map[i][j]);
        }
        ans=n*(n-1);
        int flag=Floyd();
        printf("Case %d: ", cas++);
        if(flag) printf("%d\n", ans);
        else printf("impossible\n");
    }
    return 0;
}