POJ 1860 Currency Exchange (Bellmanford algorithm inverse use to determine if there is a positive ring)

Topic links

POJ1860 Topic

There are N (1≤\len≤\le100) currencies (numbered 1 to n), m currency 22 redemption laws (including exchange rate R and handling fee C), the specific law is: When you exchange B currency in 100A, a to B exchange rate is 29.75, the handling fee is 0.39, then you can get (100- 0.39) * 29.75 = 2963.3975 b currency. Now one person has the S-model currency, which is V, and asks if he can go through the exchange and eventually get the S model currency and increase the amount (currency cannot be negative during the transaction). Analysis

A currency is regarded as a node on a graph, and a method of exchange can be regarded as an edge.
The weight of A to B is (Va-cab) *rab, note that the weight of s to S is v.
After the composition can be found that the topic is to determine whether the graph constitutes a positive right loop, so you can use Bellman-ford algorithm to find the longest road, with the opposite relaxation conditions relaxation N-1 wheel, if you can continue to relax there is a positive right loop. Code

#include <iostream> #include <cstdio> using namespace std;
const int maxm=210;

const int maxn=110;
    struct Edge {int beg;
    int end;
Double r,c;
};

Edge EDGE[MAXM];
int edgenum,n,m,s;
Double Dis[maxn],val;
    void Add_edge (int beg,int end,double r,double c) {edgenum++; edge[edgenum].beg=beg;//Start currency edge[edgenum].end=end;//End currency edge[edgenum].r=r;//exchange rate edge[edgenum].c=c;//fee} Boo L Bellmen_ford () {for (int i=1;i<=n;i++) dis[i]=0;//here is exactly the opposite of Bellman's purpose.
        Initialize the amount of money in the initial hand of the source point to the point distance of Infinity dis[s]=val;//for (int k=1;k<=n-1;k++)//N-1 wheel slack {bool Flag=false;
            for (int i=1;i<=edgenum;i++) {int U=edge[i].beg;
            int v=edge[i].end; if (dis[v]< (DIS[U]-EDGE[I].C) *EDGE[I].R)//Find the longest path {////to compare is "a point to the right of their own
                The right flag=true of the value "and" one point to another;
            dis[v]= (DIS[U]-EDGE[I].C) *EDGE[I].R; }} if (!flag) RETUrn false;
            } for (int i=1;i<=edgenum;i++)//positive ring can be infinitely relaxed if (dis[edge[i].end]< (DIS[EDGE[I].BEG]-EDGE[I].C) *EDGE[I].R)
    return true;
return false;
    } int main () {int a A, B;
    Double R_ab,c_ab,r_ba,c_ba;
        while (scanf ("%d%d%d%lf", &n,&m,&s,&val)!=eof) {edgenum=0; for (int i=1;i<=m;i++) {scanf ("%D%D%LF%LF%LF%LF",&a,&b,&r_ab,&c_ab,&r_ba,&
            C_BA);
            Add_edge (A,b,r_ab,c_ab);
        Add_edge (B,A,R_BA,C_BA);
        } bool Ans=bellmen_ford ();
        if (ans) printf ("yes\n");
    else printf ("no\n");
} return 0;
 }