POJ 1860 Currency Exchange (Bellman)

Currency Exchange
Time Limit: 1000MS
 
Memory Limit: 30000K
Total Submissions: 15324
 
Accepted: 5275
Description
Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the same pair of currencies. Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency. 
For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR. 
You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges, and real R
AB, C
AB, R
BA and C
BA - exchange rates and commissions when exchanging A to B and B to A respectively. 
Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative sum of money while making his operations. 
Input
The first line of the input contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1<=S<=N<=100, 1<=M<=100, V is real number, 0<=V<=10
3. 
For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10
-2<=rate<=10
2, 0<=commission<=10
2. 
Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations will be less than 10
4. 
Output
If Nick can increase his wealth, output YES, in other case output NO to the output file.
Sample Input

3 2 1 20.0

1 2 1.00 1.00 1.00 1.00

2 3 1.10 1.00 1.10 1.00

Sample Output

YES

Source
Northeastern Europe 2001 , Northern Subregion
 
 
제목: 모든 화폐 간에 일정한 교환 비율이 있고 일정한 커미션을 지불해야 한다.nick에게 여러 차례의 교환을 거친 후에 돈을 벌 수 있느냐고 묻는다
사고방식: 원래는 NE차 느슨함을 거친 후에 디스[src]>w만 있으면 돈을 벌 수 있을 거라고 생각했는데 생각해 보니 틀렸다.그러나 정환이 존재하는지 아닌지를 판단해야 한다. NE차 느슨해질 수 있기 때문에dis[src]는 w보다 크지 않지만 정환이 존재한다. 다시 느슨해지면dis[src]>w가 반드시 있다.
 
 

#include<iostream>

#include<cstdio>

#include<cstring>



using namespace std;



const int VM=120;

const int EM=120;

const int INF=0x3f3f3f3f;



struct Edge{

    int u,v;

    double r,c;

}edge[EM<<1];



int n,m,s,cnt;

double V,dis[VM];



void addedge(int cu,int cv,double cr,double cc){

    edge[cnt].u=cu;     edge[cnt].v=cv;     edge[cnt].r=cr;

    edge[cnt].c=cc;     cnt++;

}



int Bellman_ford(){

    int i,j;

    for(i=1;i<=n;i++)

        dis[i]=0;

    dis[s]=V;

    for(i=1;i<=n;i++){

        int flag=0;

        for(j=0;j<cnt;j++)

            if(dis[edge[j].v]<(dis[edge[j].u]-edge[j].c)*edge[j].r){

                dis[edge[j].v]=(dis[edge[j].u]-edge[j].c)*edge[j].r;

                flag=1;

            }

        if(!flag)   // 

            break;

    }

    return i==n+1;      // 

}



int main(){



    //freopen("input.txt","r",stdin);



    while(~scanf("%d%d%d%lf",&n,&m,&s,&V)){

        cnt=0;

        int u,v;

        double ruv,cuv,rvu,cvu;

        while(m--){

            scanf("%d%d%lf%lf%lf%lf",&u,&v,&ruv,&cuv,&rvu,&cvu);

            addedge(u,v,ruv,cuv);

            addedge(v,u,rvu,cvu);

        }

        int ans=Bellman_ford();

        printf("%s
",ans==1?"YES":"NO");

    }

    return 0;

}