hdu 4034 Graph（floyd）

题意：给定原图顶点间的最短路，问原图最少有几条边。

分析：若两点的距离可由另外两条边来代替，则可删去该边；若存在比这两点的距离更短的路径，则“impossible”.

/*
Author:Zhaofa Fang
Lang:C++
*/
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <utility>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
typedef long long ll;
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define lowbit(x) (x&(-x))
const int INF = 2147483647;

int maz[105][105];
int floyd(int n)
{
    int cnt = 0;
    for(int k=0;k<n;k++)
    {
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(i == k || j == k || i == k)continue;
                if(maz[i][k] == -1 || maz[k][j] == -1 || maz[i][j] == -1)continue;
                if(maz[i][j] > maz[i][k] + maz[k][j])return -1;
                if(maz[i][j] == maz[i][k] + maz[k][j])
                {
                    cnt++;maz[i][j]=-1;
                }
            }
        }
    }
    return n*(n-1) - cnt ;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in","r",stdin);
    #endif
    int T,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        memset(maz,0,sizeof(maz));
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                scanf("%d",&maz[i][j]);
        int tmp = floyd(n);
        printf("Case %d: ",++cas);
        if(tmp != -1)printf("%d\n",tmp);
        else puts("impossible");
    }
    return 0;
}