poj1860解题报告

来源：互联网发布：民谣知乎编辑：程序博客网时间：2024/05/23 01:23

题目大意为：存在n种货币，现在拥有一定数量的某种货币，各种货币之间可以在支付一定的手续费后相互兑换，问是否存在一种兑换方法可以使拥有的货币量不断增值。

解题思路：通过考虑Bellman-Ford算法来构造图的“最短路径”，即通过松弛来进行货币的可行兑换方案，最后检测是否存在正权回路，从而得出答案。

//POJ1860-Currency Exchange//带负值的最短路径求解，Bellman-Ford算法//212K32MS#include <iostream>using namespace std;//#define IniDis   -88888.0     //初始各个点到起始点的距离struct Edge             //边的结构体，包括起始点和目的点以及汇率和手续费{int from;int to;double rate;double commission;};double Distance[110];            //各种货币可以换取到的最大数目Edge pEdgeInfo[202];int Bellman_Ford(int node, int edge, int startnode, double iniMoney){double temp  = 0;int nflag = 0;int nchange = 0;//for (int i = 0; i < node+1; i++)//{//Distance[i] = IniDis;   //源点到各个目的点的距离不能初始化为负数，否则会WA//}memset(Distance,0,sizeof(Distance));Distance[startnode] = iniMoney;for (int i = 1; i< node; i++){nchange = 0;for (int j = 0; j< edge; j++){temp = (Distance[pEdgeInfo[j].from] - pEdgeInfo[j].commission )*pEdgeInfo[j].rate;   //松弛，找到更好的兑换方式if (Distance[pEdgeInfo[j].from] > 0 && temp > Distance[pEdgeInfo[j].to]){Distance[pEdgeInfo[j].to] = temp;nchange = -1;}}if (nchange == 0){break;}}for (int j = 0; j< edge; j++)        //检测是否有正回路{temp = (Distance[pEdgeInfo[j].from] - pEdgeInfo[j].commission )*pEdgeInfo[j].rate;   //松弛，找到更好的兑换方式if (temp > Distance[pEdgeInfo[j].to]){nflag = -1;break;}}return nflag;}int main(void){int nCurrency, excPoint, curStart;        //货币总样数，交互点数目,初始货币种类double curTotal;                                 //初始货币金额int eNum;int a,b;double rab,cab,rba,cba;int nflag;                                   //检测算法的返回情况while(cin>>nCurrency>>excPoint>>curStart>>curTotal){nflag = 0;eNum =0;for (int i = 1; i< excPoint+1; i++)                           //初始化交互点数据，即图中的各条边信息{cin>>a>>b>>rab>>cab>>rba>>cba;pEdgeInfo[eNum].from = a;pEdgeInfo[eNum].to = b;pEdgeInfo[eNum].rate = rab;pEdgeInfo[eNum].commission = cab;eNum++;pEdgeInfo[eNum].from = b;pEdgeInfo[eNum].to = a;pEdgeInfo[eNum].rate = rba;pEdgeInfo[eNum].commission = cba;eNum++;}nflag = Bellman_Ford(nCurrency, 2*excPoint, curStart, curTotal);if (nflag == -1)cout<<"YES"<<endl;elsecout<<"NO"<<endl;}return 0;}

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