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。。。。。
1/* ***********************************************
2Author :111qqz
3Created Time :2016年05月24日 星期二 23时41分46秒
4File Name :code/poj/1860.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=1E2+7;
34int n,m,s;
35double v;
36double d[N];
37bool inq[N];
38vector <pair<int,pair<double,double> > > edge[N];
39
40int dblcmp(double d)
41{
42 return d<-eps?-1:d>eps;
43}
44bool spfa(int s,double amt)
45{
46 ms(d,0);
47 ms(inq,false);
48 queue<int>q;
49 q.push(s);
50 inq[s] = true;
51 d[s] = amt;
52
53 while (!q.empty())
54 {
55 int u = q.front();
56 q.pop();
57 inq[u] = false;
58
59 int siz = edge[u].size();
60 for ( int i = 0 ; i < siz; i ++)
61 {
62 int v = edge[u][i].fst;
63 double r = edge[u][i].sec.fst;
64 double c = edge[u][i].sec.sec;
65 double tmp = (d[u]-c)*r;
66 if (dblcmp(d[v]-tmp)<0)
67 {
68 d[v] = tmp;
69 if (inq[v]) continue;
70 inq[v] = true;
71 q.push(v);
72 }
73 }
74 }
75 return d[s]>amt;
76}
77int main()
78{
79 #ifndef ONLINE_JUDGE
80 freopen("code/in.txt","r",stdin);
81 #endif
82
83 ios::sync_with_stdio(false);
84 cin>>n>>m>>s>>v;
85 for ( int i = 1 ; i <= m ; i++)
86 {
87 int u,v;
88 double r1,c1,r2,c2;
89 cin>>u>>v>>r1>>c1>>r2>>c2;
90 edge[u].push_back(MP(v,MP(r1,c1)));
91 edge[v].push_back(MP(u,MP(r2,c2)));
92 }
93
94 if (spfa(s,v))
95 {
96 puts("YES");
97 }
98 else
99 {
100 puts("NO");
101 }
102 #ifndef ONLINE_JUDGE
103 fclose(stdin);
104 #endif
105 return 0;
106}