Light OJ 1221 – Travel Company (Floyed Warshal)

1

Problem Link : http://www.lightoj.com/volume_showproblem.php?problem=1221



/*
         +-+ +-+ +-+
         |R| |.| |S|
         +-+ +-+ +-+
 */

#include <bits/stdc++.h>

#define pii             pair <int,int>
#define sc              scanf
#define pf              printf
#define Pi              2*acos(0.0)
#define ms(a,b)         memset(a, b, sizeof(a))
#define pb(a)           push_back(a)
#define MP              make_pair
#define db              double
#define ll              long long
#define EPS             10E-10
#define ff              first
#define ss              second
#define sqr(x)          (x)*(x)
#define D(x)            cout<<#x " = "<<(x)<<endl
#define VI              vector <int>
#define DBG             pf("Hi\n")
#define MOD             100007
#define MAX             105
#define CIN             ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)           (int)a.size()
#define sf(a)           scanf("%d",&a)
#define sff(a,b)        scanf("%d%d",&a,&b)
#define sfff(a,b,c)     scanf("%d%d%d",&a,&b,&c)
#define loop(i,n)       for(int i=0;i<n;i++)
#define REP(i,a,b)      for(int i=a;i<b;i++)
#define TEST_CASE(t)    for(int z=1;z<=t;z++)
#define PRINT_CASE      printf("Case %d: ",z)
#define all(a)          a.begin(),a.end()
#define intlim          2147483648
#define inf             1000000
#define ull             unsigned long long

using namespace std;

int graph[MAX][MAX];

void allclear(int n)
{
    for(int i=0; i<n+2; i++)
    {
        for(int j=0; j<=n+2; j++)
        {
            graph[i][j]=inf;
        }
        graph[i][i]=0;
    }
}

int main()
{
    ///freopen("in.txt","r",stdin);
    ///freopen("out.txt","w",stdout);
    int t;
    sf(t);
    TEST_CASE(t)
    {
        int n,r,p;
        sfff(n,r,p);
        allclear(n);
        int a,b,income,expence;
        loop(i,r)
        {
            sc("%d %d %d %d",&a,&b,&income,&expence);
            graph[a][b]=p*expence-income;
        }

        REP(k,0,n) REP(i,0,n) REP(j,0,n) graph[i][j]=min(graph[i][j],graph[i][k]+graph[k][j]);
        bool negative_cycle=0;
        REP(i,0,n)
        if(graph[i][i]<0)
            negative_cycle=1;
        PRINT_CASE;
        if(negative_cycle)
            pf("YES\n");
        else
            pf("NO\n");
    }
    return 0;
}

Another Solution Using Bellman Ford is here.. -> Light OJ 1221 – Travel Company (Bellman Ford)