hdu4034 Graph(floyd)

题目：  给出一个图的最短路，求原图最少几条边

Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1695    Accepted Submission(s): 848

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

 

 

 

先假设每两个点之间都有边，ans = n*(n-1), 然后用floyd ， 对于 i,j , 以k为中点，如果 dist[i][j] == dist[i][k]+dist[k][j] 说明这条路可以被替代，减去一条边。

反之如果大于的话，就和题目的最短路矛盾了。 输出impossible

 

代码：

#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>
using namespace std;
#define XINF INT_MAX
#define INF 0x3FFFFFFF
#define MP(X,Y) make_pair(X,Y)
#define PB(X) push_back(X)
#define REP(X,N) for(int X=0;X<N;X++)
#define REP2(X,L,R) for(int X=L;X<=R;X++)
#define DEP(X,R,L) for(int X=R;X>=L;X--)
#define CLR(A,X) memset(A,X,sizeof(A))
#define IT iterator
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<PII> VII;
typedef vector<int> VI;

int N;
int dist[105][105];
int vis[105][105];
int main()
{
    ios::sync_with_stdio(false);

    int tst;
    cin>>tst;
    int cse = 0;
    while(tst--)
    {
        CLR(dist,0);CLR(vis,0);

        cin>>N;
        REP(i,N)
        REP(j,N)
        {
            cin>>dist[i][j];
        }
        cout<<"Case "<<++cse<<": ";

        int ans = N*(N-1);
        REP(k,N)
        {
            REP(i,N)
            {
                REP(j,N)
                {
                    if( i == k || j == k)continue;
                    if(!vis[i][j] && dist[i][j] == dist[i][k]+dist[k][j])
                    {
                        ans--;
                        vis[i][j] =1;
                    }
                    if( dist[i][j] > dist[i][k]+dist[k][j])
                    {
                        ans = -1;
                        break;
                    }
                }
                if( ans ==-1)
                {
                    break;
                }
            }
            if(ans==-1)break;
        }
        if( ans == -1)
        {
            cout<<"impossible"<<endl;
        }
        else
        {
            cout<<ans<<endl;
        }

    }

    return 0;
}