POJ1860 Currency Exchange【BellmanFord算法】【求正权回路】

Currency Exchange

Time Limit: 1000MSMemory Limit: 30000K

Total Submissions: 20994Accepted: 7522

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

Sample Output

YES

Source

Northeastern Europe 2001, Northern Subregion

题目大意：有N种货币，货币之间可以按汇率交换，同时还需要收手续费，当你用100A货币去交换B货币，

假如A到B的汇率为29.75，手续费为0.39，则你可以得到(100-0.39)*29.75 = 2963.3975的B货币。货币

可以一直重复交换，问：能否通过兑换货币之后，增加你手中货币的价值，则输出"YES"，否则输出"NO"。

思路：把N种货币看成图上的N个点，当你有数量为V的货币A时，

货币AB之间的权值就是——(V-手续费)*A到B的汇率

这道题就可以转换为求图是否还有可无限增大(含有正权回路)的最大路径。那么怎么来判断是否含有正权回

路。可以用BellmanFord算法的思想来做。

BellmanFord是来求最短路径，并判断是否存在负权回路。这里用BellmanFord来求最长路径，并判断是否

存在正权回路。

<span style="font-family:Microsoft YaHei;">#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;const int MAXN = 110;const int MAXM = 220;const int INF = 0xffffff0;struct EdgeNode{    int to;     //连接的b点    int next;   //    double rate;//    double cost;}Edges[MAXM];int Head[MAXN];double Dist[MAXN];//M——兑换点个数 N——货币种数 S——手里持有第S种货币 V——持有的第S种货币资金bool BellmanFord(int M,int N,int S,double V){    Dist[S] = V;    for(int i = 1; i < N; ++i)    {        for(int j = 1; j <= N; ++j)        {            for(int k = Head[j]; k != -1; k = Edges[k].next)    //寻找最长路径            {                if(Dist[Edges[k].to] < (Dist[j]-Edges[k].cost)*Edges[k].rate)                    Dist[Edges[k].to] = (Dist[j]-Edges[k].cost)*Edges[k].rate;            }        }    }    for(int j = 1; j <= N; ++j)    {        for(int k = Head[j]; k != -1; k = Edges[k].next)        {            if(Dist[Edges[k].to] < (Dist[j]-Edges[k].cost)*Edges[k].rate)   //S到其他路径能一直增大时，说明存在最长路径                return true;        }    }    return false;}int main(){    int N,M,S,u,v;    double V,Rab,Cab,Rba,Cba;    while(~scanf("%d%d%d%lf",&N,&M,&S,&V))    {        int id = 0;        memset(Head,-1,sizeof(Head));        memset(Dist,0,sizeof(Dist));            //初始为0，不为INF，因为BellmanFord是用来找负环的，这里用来找是否存在正环        memset(Edges,0,sizeof(Edges));        for(int i = 0; i < M; ++i)        {            scanf("%d%d%lf%lf%lf%lf",&u,&v,&Rab,&Cab,&Rba,&Cba);            Edges[id].to = v;            Edges[id].rate = Rab;            Edges[id].cost = Cab;            Edges[id].next = Head[u];            Head[u] = id++;            Edges[id].to = u;            Edges[id].rate = Rba;            Edges[id].cost = Cba;            Edges[id].next = Head[v];            Head[v] = id++;        }        if(BellmanFord(id,N,S,V))            printf("YES\n");        else            printf("NO\n");    }    return 0;}</span>