poj1860(求是否存在大于0的路径,spfa算法) Currency Exchange

2017-07-12 90点热度 0人点赞
Currency Exchange
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 30362   Accepted: 11474

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 RAB, CAB, RBA and CBA - 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<=103
For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10-2<=rate<=102, 0<=commission<=102
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 104

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

//求是否存在大于0的路径,spfa算法,
//如果一个点进队列n次,则必定一个正环,将可以一直迭代
//我发现spfa和Bellman好像是一个东西
#include
#include
#include
#include
using namespace std;
int n,m,s,head[210],inq[210];
int cnt[210];
double vv,dis[210];
struct node
{
    int to,next;
    double r,c;
} edge[300];
bool bellman()
{
    for(int i=1; i<=n; i++) dis[i]=-1e+18;
    memset(inq,0,sizeof(inq));
    memset(cnt,0,sizeof(cnt));
    queue q;
    q.push(s);
    dis[s]=vv;
    inq[s]=1,cnt[s]++;
    while(!q.empty())
    {
        int x=q.front();
        q.pop();
        inq[x]=0;//这个不要忘了
        for(int i=head[x]; i!=-1; i=edge[i].next)
        {
            int d=edge[i].to;
            if(dis[d]<(dis[x]-edge[i].c)*edge[i].r)
            {
                dis[d]=(dis[x]-edge[i].c)*edge[i].r;
                if(!inq[d])
                {
                    q.push(d);
                    inq[d]=1;
                    cnt[d]++;
                    if(cnt[d]>=n) return true;
                }
            }
        }
    }
    return false;
}
int main()
{
    ios::sync_with_stdio(0);
    int u,v;
    double r1,c1,r2,c2;
    while(cin>>n>>m>>s>>vv)
    {
        memset(head,-1,sizeof(head));
        for(int i=0; i>u>>v>>r1>>c1>>r2>>c2;
            edge[i].to=v;
            edge[i].r=r1;
            edge[i].c=c1;
            edge[i].next=head[u];
            head[u]=i;

            edge[i+m].to=u;
            edge[i+m].r=r2;
            edge[i+m].c=c2;
            edge[i+m].next=head[v];
            head[v]=i+m;
        }
        if(bellman()) cout<<"YESn";
        else cout<<"NOn";
    }
    return 0;
}

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