poj 1860 Currency Exchange (spfa求最长路)
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
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=1E2+7;
7int n,m,s;
8double v;
9double d[N];
10bool inq[N];
11vector <pair<int,pair<double,double> > > edge[N];
1int dblcmp(double d)
2{
3 return d<-eps?-1:d>eps;
4}
5bool spfa(int s,double amt)
6{
7 ms(d,0);
8 ms(inq,false);
9 queue<int>q;
10 q.push(s);
11 inq[s] = true;
12 d[s] = amt;
1 while (!q.empty())
2 {
3 int u = q.front();
4 q.pop();
5 inq[u] = false;
1 int siz = edge[u].size();
2 for ( int i = 0 ; i < siz; i ++)
3 {
4 int v = edge[u][i].fst;
5 double r = edge[u][i].sec.fst;
6 double c = edge[u][i].sec.sec;
7 double tmp = (d[u]-c)*r;
8 if (dblcmp(d[v]-tmp)<0)
9 {
10 d[v] = tmp;
11 if (inq[v]) continue;
12 inq[v] = true;
13 q.push(v);
14 }
15 }
16 }
17 return d[s]>amt;
18}
19int main()
20{
21 #ifndef ONLINE_JUDGE
22 freopen("code/in.txt","r",stdin);
23 #endif
1 ios::sync_with_stdio(false);
2 cin>>n>>m>>s>>v;
3 for ( int i = 1 ; i <= m ; i++)
4 {
5 int u,v;
6 double r1,c1,r2,c2;
7 cin>>u>>v>>r1>>c1>>r2>>c2;
8 edge[u].push_back(MP(v,MP(r1,c1)));
9 edge[v].push_back(MP(u,MP(r2,c2)));
10 }
1 if (spfa(s,v))
2 {
3 puts("YES");
4 }
5 else
6 {
7 puts("NO");
8 }
9 #ifndef ONLINE_JUDGE
10 fclose(stdin);
11 #endif
12 return 0;
13}