Currency Exchange
Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u
Submit Status Practice POJ 1860
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, CAB, 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：给你n种货币，m种货币交换关系，交换率和手续费，给你起始的货币类型和金额，问你是否可以通过交换货币，最后回到起始的货币时能盈利。 解题思路：如果要盈利，只需要判断图中存不存在正环， 即可以一直让某种货币额度无限增加。由于是无向图，那么只要存在正环，那么我就可以最后转化成起始的货币且盈利。所以只要将SPFA判负环的条件变化一下就行。初始值时，让除原点之外的d数组都赋值为0。同时松弛条件变为d[e.to] < (d[e.from] - e.com)*e.rate。即可，最后判断当u为起点时的d[u]是否大于起始金额即可。
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<queue>
#include<vector>
#include<iostream>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1e3+200;
int n , m;
struct Edge{
int from,to;
double rate , com;
};
vector<Edge>edges;
vector<int>G[maxn];
void init(){
for(int i = 0; i <= n; i++){
G[i].clear();
}
edges.clear();
}
double d[maxn] ,cnt[maxn], inq[maxn];
void AddEdge(int u,int v,double r,double co){
edges.push_back( (Edge){u,v,r,co} );
m = edges.size();
G[u].push_back(m-1);
}
bool SPFA(int s, double V){
queue<int>Q;
for(int i = 0; i <= n; i++){
d[i] = 0;
}
d[s] = V;
cnt[s] ++;
inq[s] = 1;
Q.push(s);
while(!Q.empty()){
int u = Q.front();
Q.pop();
if(u == s&& d[s] > V){
return true;
}
inq[u] = 0;
for(int i = 0; i < G[u].size(); i++){
Edge & e = edges[G[u][i]];
if(d[e.to] < (d[e.from] - e.com)*e.rate ){
d[e.to ] = (d[e.from] - e.com) *e.rate;
if(!inq[e.to]){
inq[e.to] = 1;
Q.push(e.to);
}
}
}
}
return false;
}
int main(){
int mm,s;
double k;
while(scanf("%d%d%d%lf",&n,&mm,&s,&k)!=EOF){
int a,b;
double c,d;
for(int i = 0; i < mm; i++){
scanf("%d%d%lf%lf",&a,&b,&c,&d);
AddEdge(a,b,c,d);
scanf("%lf%lf",&c,&d);
AddEdge(b,a,c,d);
}
bool yes = SPFA(s,k);
if(yes){
puts("YES");
}else{
puts("NO");
}
}
return 0;
}