Graph

梦里梦外; 2022-06-16 01:07 158阅读 0赞

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 ][The 36th ACM_ICPC Asia Regional Chengdu Site _ Online Contest]

Recommend

lcy   |   We have carefully selected several similar problems for you:   [4039][]  [4038][]  [4036][]  [4033][]  [4037][]

#### 题目大意：给出一张图，表示的是点与点之间的最短路，问原图最少有几条边。不存在这种情况就输出 impossible ####

#include <cstdio>
    #include <cstring>
    #include <string>
    #include <iostream>
    #include <algorithm>
    using namespace std;
    #define INF 0xffffff
    int n,coun,i,j,k;
    int map[147][147];
    int Floyd()
    {
        for(i = 1 ; i <= n ; i++)
        {
            for(j = 1 ; j <= n ; j++)
            {
                if(map[i][j] == 0)
                    continue;
                for(k = 1 ; k <= n ; k++)
                {
                    if( map[i][k] != 0 && map[k][j] != 0)
                    {
                        if(map[i][j] > map[i][k] + map[k][j])
                            return 0;
                        else if(map[i][j] == map[i][k] + map[k][j])
                        {
                            coun--;
                            break;//因为有总共n*(n-1)条路径,即（i,j）,其中i!=i;
                            //也就是（1,2）（1,3）（1,4）...(2,1)(2,3)(2,4)....
                            //(3,1)(3,2)(3,4)...(4,1)(4,2)(4,3).....
                            //所以当map[i][j] == map[i][k] + map[k][j]时，必须跳出循环，相当于i到j（直达）这条路已经不存在了
                            //既然不存在了就不用再去减了，当然你要减去也行（即不跳出循环），但是你得把它标记一下，说明它已经不存在了，只有当存在的时候才减
                        }
                    }
                }
            }
        }
        return coun;
    }
    int main()
    {
        int T,cas;
        scanf("%d",&T);
        cas = 0;
        while(T--)
        {
            scanf("%d",&n);
            for(i = 1; i <= n ; i++)
                for(j = 1 ; j <= n ; j++)
                    scanf("%d",&map[i][j]);
            coun=n*(n-1);
            int ans = Floyd();
            if(ans != 0)
                printf("Case %d: %d\n",++cas,ans);
            else
                printf("Case %d: impossible\n",++cas);
        }
        return 0;
    }

[The 36th ACM_ICPC Asia Regional Chengdu Site _ Online Contest]: http://acm.hdu.edu.cn/search.php?field=problem&key=The+36th+ACM%2FICPC+Asia+Regional+Chengdu+Site+%A1%AA%A1%AA+Online+Contest&source=1&searchmode=source
[4039]: http://acm.hdu.edu.cn/showproblem.php?pid=4039
[4038]: http://acm.hdu.edu.cn/showproblem.php?pid=4038
[4036]: http://acm.hdu.edu.cn/showproblem.php?pid=4036
[4033]: http://acm.hdu.edu.cn/showproblem.php?pid=4033
[4037]: http://acm.hdu.edu.cn/showproblem.php?pid=4037