Graph
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 962 Accepted Submission(s): 510
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.
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
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lcy
#include<stdio.h> const int MAXN=110; int map[MAXN][MAXN]; int ans;//至少需要的边数 int n;//点数 int Solve() { for(int i=0;i<n;i++) for(int j=0;j<n;j++) for(int k=0;k<n;k++) { if(i!=j&&j!=k&&i!=k) { if(map[i][j]==map[i][k]+map[k][j]) { ans--;//这条边可以不需要的 break; } if(map[i][j]>map[i][k]+map[k][j]) return 0;//不可能的情况 } } return 1; } int main() { int T; scanf("%d",&T); while(T--) { 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);//最多的边数 if(Solve()) { printf("%d\n",ans); } else printf("impossible\n"); } return 0; }