（floyd） hdu 4034

Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 2030    Accepted Submission(s): 1014

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

 

给出一个方阵，a[i][j]表示i到j的最短路

求最少需要多少边的图符合这个方阵

 

解析：

 

直接floyd 啊

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
int n,mp[101][101];
bool vis[101][101],flag;
void floyd()
{
    for(int k=1;k<=n;k++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(k==i||i==j||j==k)
                    continue;
                if(mp[i][k]+mp[k][j]==mp[i][j])
                    vis[i][j]=0;
                if(mp[i][k]+mp[k][j]<mp[i][j])
                {
                    flag=false;
                    break;
                }
            }
            if(!flag)
                break;
        }
        if(!flag)
            break;
    }
}
int main()
{
    int tt,cas=1;
    scanf("%d",&tt);
    while(tt--)
    {
        int ans=0;
        scanf("%d",&n);
        flag=true;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
                scanf("%d",&mp[i][j]),vis[i][j]=1;
        }
        floyd();
        if(!flag)
        {
            printf("Case %d: impossible\n",cas++);
            continue;
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(mp[i][j]==0)
                    vis[i][j]=0;
            }
        }
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                if(vis[i][j])
                    ans++;
            }
        }
        printf("Case %d: %d\n",cas++,ans);
    }
    return 0;
}