I - Currency Exchange

题目:
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

代码如下:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define MAX 105
struct Node
{
    int x,y;
    double rate,cost;
}node[500];
int n,m,s,x1,y1,k;
double dis[MAX],v;//dis表示从开始的货币一直兑换到第i种货币还剩多少钱
bool Bellmam_Ford()
{
    bool flag;
    memset(dis,0,sizeof(0));//这里与普通的Bellmam-Ford算法不同，初始状况每一种货币都为0
    dis[s] = v;
    for(int i = 1;i <= n - 1;i++){
        flag = false;
        for(int j = 0;j < k;j++){//如果兑换一种货币使得钱数量增加则代表是可以松弛的
            if(dis[node[j].y] < (dis[node[j].x] - node[j].cost) * node[j].rate){
                dis[node[j].y] = (dis[node[j].x] - node[j].cost) * node[j].rate;
                flag = true;//松弛成功
            }
        }
        if(!flag) break;
    }
    for(int j = 0;j < k;j++){
        if(dis[node[j].y] < (dis[node[j].x] - node[j].cost) * node[j].rate)
            return true;
    }
    return false;
}
int main()
{
    double rate1,rate2,cost1,cost2;
    while(~scanf("%d%d%d%lf",&n,&m,&s,&v)){
        k = 0;
        for(int i = 0;i < m;i++){
            scanf("%d%d%lf%lf%lf%lf",&x1,&y1,&rate1,&cost1,&rate2,&cost2);
            node[k].x = x1;
            node[k].y = y1;
            node[k].rate = rate1;
            node[k++].cost = cost1;
            node[k].x = y1;
            node[k].y = x1;
            node[k].rate = rate2;
            node[k++].cost = cost2;
        }
        if(Bellmam_Ford()) cout << "YES" << endl;
        else cout << "NO" << endl;
    }
    return 0;
}

这里每一种货币都可以看成一个点，每一种货币兑换成另外一种货币就是两点之间的权值。现在题目要求求s币的金额经过交换最终得到的s币金额数能否增加，换而言之如果图中出现正权回路那么就输出YES，否则就是NO。
本题可以使用Bellmam-Ford算法，虽然Bellmam-Ford算法是用来处理负权边以及处理负环问题的，但是这要将Bellmam-Ford算法的松弛条件改反一下，就可以解决这个问题。如果经过n - 1次松弛之后，还能再次松弛成功那就输出YES，否则就是NO。