POJ-1860 Currency Exchange

Currency Exchange

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<=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 20.01 2 1.00 1.00 1.00 1.002 3 1.10 1.00 1.10 1.00

Sample Output

YES

————————————————————集训14.1的分割线————————————————————

思路：这题和POJ-2240 Arbitrage几乎是一样的。多了一个手续费。跑一次SPFA找最长路径，判断“增权值回路”即可。挺水的。

P.S. 由于小明手中的钱限制，必须跑他有的那种纸币才行，不能任选了。

代码如下：

/*ID: j.sure.1PROG:LANG: C++*//****************************************/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <string>#include <climits>#include <iostream>#define INF 0x3f3f3f3fusing namespace std;/****************************************/const int N = 105, M = 105;int n, m, s, tot, head[N], enq[N], q[N];double dis[N], money;bool inq[N];struct Node{    int u, v, next;double rate, tax;}edge[2*M];void init(){    memset(head, -1, sizeof(head));    tot = 0;}void add(int u, int v, double rate, double tax){    edge[tot].u = u; edge[tot].v = v;edge[tot].rate = rate; edge[tot].tax = tax;    edge[tot].next = head[u]; head[u] = tot++;}bool spfa(int st){    for(int i = 1; i <= n; i++) {inq[i] = enq[i] = dis[i] = 0;}int fron = 0, rear = 1;q[fron] = st;dis[st] = money;    inq[st] = true;enq[st]++;    while(fron < rear) {        int u = q[fron%N]; fron++;        inq[u] = false;        for(int i = head[u]; i != -1; i = edge[i].next) {            int v = edge[i].v;            if(dis[v] < (dis[u] - edge[i].tax)*edge[i].rate) {                dis[v] = (dis[u] - edge[i].tax)*edge[i].rate;                if(!inq[v]) {                    q[rear%N] = v; rear++;enq[v]++;if(enq[v] > n) return true;inq[v] = true;                }            }        }    }return false;}int main(){#ifdef J_Surefreopen("000.in", "r", stdin);//freopen(".out", "w", stdout);#endif    scanf("%d%d%d%lf", &n, &m, &s, &money);init();int u, v;double rate_ab, tax_ab, rate_ba, tax_ba;for(int i = 0; i < m; i++) {scanf("%d%d%lf%lf%lf%lf", &u, &v, &rate_ab, &tax_ab, &rate_ba, &tax_ba);add(u, v, rate_ab, tax_ab);add(v, u, rate_ba, tax_ba);}if(spfa(s)) puts("YES");else puts("NO");    return 0;}