POJ-1860 Currency Exchange（Bellman-Ford）

Currency Exchange

http://poj.org/problem?id=1860

Time Limit: 1000MS		Memory Limit: 30000K

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

题目大意：有n种货币，2*m种兑换关系，每次兑换需要支付c元手续费，汇率为r，初始有s货币v元，求是否能通过兑换，使持有的s货币变多？

由于本题是要求走一圈后，权值变大，所以更改Bellman-Ford的初始条件和更新条件，即可更改为求已s为起点的“最长路”，因为更新方法变了，所以每次能更新到的点必定是s点可达的点，即也再可兑换回s，所以只要用Bellman-Ford判断是否有“负权回路”即可

关于讨论版后的数据：

45 43 8 32.460000
40 6 90.90 17.59 45.28 89.70
29 18 37.97 55.13 84.48 52.81
45 26 59.25 99.43 27.79 90.84
7 42 57.94 54.76 90.96 64.60
27 12 18.69 31.72 66.78 65.72          //5
31 4 48.09 14.86 56.63 18.38
23 1 96.51 39.40 14.79 23.51
22 25 85.68 78.28 77.54 0.17
18 5 28.25 56.09 2.81 25.39
34 42 91.34 9.29 34.51 99.34
20 32 39.35 63.72 58.26 19.53
36 34 1.79 61.68 23.29 4.84
30 12 70.83 5.46 4.05 31.37           //13
16 37 39.47 57.44 59.00 79.05
16 14 68.12 5.58 35.39 19.28
13 44 27.08 32.10 75.83 47.52
45 37 6.31 81.34 22.96 54.24
24 19 41.55 58.31 93.99 84.14
28 36 73.32 80.70 37.02 33.93
6 21 56.88 6.52 76.14 30.43
18 35 83.55 30.02 41.25 89.11
17 37 70.84 51.14 44.76 29.99
14 2 42.34 26.79 51.95 93.56
40 6 63.46 83.44 82.66 15.09
27 20 34.61 61.69 10.23 35.04
29 13 83.62 38.65 88.51 27.75
30 22 59.09 3.86 63.35 57.85
4 18 11.24 97.52 8.80 11.15
38 42 79.71 33.62 3.36 36.53
38 45 90.65 72.52 18.35 54.58
33 45 16.35 58.44 57.71 88.69
1 43 84.41 35.96 72.06 27.57
13 9 46.36 94.47 81.63 88.19
42 10 41.22 1.63 98.33 85.94
9 3 40.00 49.78 46.31 86.28
11 37 27.72 17.84 56.89 86.23
45 10 99.57 92.25 71.49 1.57
8 14 19.47 82.06 32.79 90.53
23 37 26.54 38.20 99.18 86.79
10 42 98.34 36.13 88.52 98.24
13 30 85.81 43.84 54.64 33.83
31 8 53.55 51.02 46.20 87.65
30 8 92.31 36.44 91.82 23.12     //43

这组数据确实应该输出YES，但是用Bellman-Ford的话，得用long double才行，否则会在更新几次之后，所有的dis全部变成inf，导致最终结果出错

但是用long double会WA，估计标程并未考虑double会溢出，所以可能数据出错了...

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int MAXN=105;
const int INF=0x3f3f3f3f;

struct Edge {
    int s,e;
    double r,c;

    Edge(int ss=0,int ee=0,double rr=0,double cc=0):s(ss),e(ee),r(rr),c(cc) {}
};

int n,m,a,b;
double s,v,r1,c1,r2,c2;
vector<Edge> edge;
double dis[MAXN];

bool Bellman_Ford(int sta) {//可判断负权回路
    bool relaxed;
    memset(dis,0,sizeof(dis));
    dis[sta]=v;//初始化sta货币有v，其余均为0

    for(int i=1;i<n;++i) {
        relaxed=false;
        for(int j=0;j<edge.size();++j) {
            if((dis[edge[j].s]-edge[j].c)*edge[j].r>dis[edge[j].e]) {//松弛，只有兑换后当前币种的货币数量上升时才更新
                dis[edge[j].e]=(dis[edge[j].s]-edge[j].c)*edge[j].r;
                relaxed=true;
            }
        }
        if(!relaxed) {//如果未更新，则不会再更新，且无负权回路
            return false;
        }
    }
    for(int j=0;j<edge.size();++j) {
        if((dis[edge[j].s]-edge[j].c)*edge[j].r>dis[edge[j].e]) {//如果可以继续松弛，则存在负权回路
            return true;
        }
    }
    return false;
}

int main() {
    while(4==scanf("%d%d%lf%lf",&n,&m,&s,&v)) {
        edge.clear();
        while(m-->0) {
            scanf("%d%d%lf%lf%lf%lf",&a,&b,&r1,&c1,&r2,&c2);
            edge.push_back(Edge(a,b,r1,c1));
            edge.push_back(Edge(b,a,r2,c2));
        }
        printf("%s\n",Bellman_Ford(s)?"YES":"NO");
    }
    return 0;
}