poj 1860 Currency Exchange (spfa求最长路)

poj 1860 题目链接

题意：有n种货币，m个货币交易点，每个货币交易点只能是两种货币之间交换，给出两个方向的汇率和手续费。初始拥有数量v的货币s,问能否经过一些py交易，使得最后手里的货币s比v多。

思路：大概还是用spfa求最长路。。松弛那里需要注意一下算法。。。

1A。。。好爽啊。。。。。

/* ***********************************************
Author :111qqz
Created Time :2016年05月24日 星期二 23时41分46秒
File Name :code/poj/1860.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=1E2+7;
int n,m,s;
double v;
double d[N];
bool inq[N];
vector <pair<int,pair<double,double> > > edge[N];

int dblcmp(double d)
{
    return d<-eps?-1:d>eps;
}
bool spfa(int s,double amt)
{
    ms(d,0);
    ms(inq,false);
    queue<int>q;
    q.push(s);
    inq[s] = true;
    d[s] = amt;

    while (!q.empty())
    {
	int u = q.front();
	q.pop();
	inq[u] = false;

	int siz = edge[u].size();
	for  ( int i = 0 ; i  < siz; i ++)
	{
	    int v = edge[u][i].fst;
	    double r = edge[u][i].sec.fst;
	    double c = edge[u][i].sec.sec;
	    double tmp = (d[u]-c)*r;
	    if (dblcmp(d[v]-tmp)<0)
	    {
		d[v] = tmp;
		if (inq[v]) continue;
		inq[v] = true;
		q.push(v);
	    }
	}
    }
    return d[s]>amt;
}
int main()
{
	#ifndef  ONLINE_JUDGE 
	freopen("code/in.txt","r",stdin);
  #endif

	ios::sync_with_stdio(false);
	cin>>n>>m>>s>>v;
	for ( int i = 1 ; i <= m ; i++)
	{
	    int u,v;
	    double r1,c1,r2,c2;
	    cin>>u>>v>>r1>>c1>>r2>>c2;
	    edge[u].push_back(MP(v,MP(r1,c1)));
	    edge[v].push_back(MP(u,MP(r2,c2)));
	}

	if (spfa(s,v)) 
	{
	    puts("YES");
	}
	else
	{
	    puts("NO");
	}
  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}