给出 N 种货币 M 条兑换关系 开始时所有的货币S 和有X 块钱

接下来M条关系

A B W1 W2 W3 W4

表示

A->B 所需的手续费为W2块钱  汇率为W1

B->A 所需的手续费为W4块钱  汇率为W3

所以对于输入的一行建两条边

要求到最后可以赚到钱

所以当出现了负圈即可赚到无限多的钱

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <cctype>
#include <cmath>
#include <string>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <iomanip>
using namespace std;
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <set>
#include <map>
typedef long long LL;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define pi acos(-1.0)
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
typedef pair<int, int> PI;
typedef pair<int, PI> PP;
#ifdef _WIN32
#define LLD "%I64d"
#else
#define LLD "%lld"
#endif
//LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}
//inline int read(){char ch=' ';int ans=0;while(ch<'0' || ch>'9')ch=getchar();while(ch<='9' && ch>='0'){ans=ans*10+ch-'0';ch=getchar();}return ans;}
//inline void print(LL x){printf(LLD, x);puts("");}
//inline void read(int &x){char c = getchar();while(c < '0') c = getchar();x = c - '0'; c = getchar();while(c >= '0'){x = x * 10 + (c - '0'); c = getchar();}}
const int INF=0x3f3f3f3f;
const int MAXN=550;
double dist[MAXN];
struct Edge
{int u,v;double a,b;Edge(int _u=0,int _v=0,double _a=0,double _b=0):u(_u),v(_v),a(_a),b(_b) {}
};
vector<Edge>E;
bool bellman_ford(int start,int n,double x)
{for(int i=1; i<=n; i++)dist[i]=0;dist[start]=x;for(int i=1; i<n; i++){bool flag=false;for(int j=0; j<E.size(); j++){int u=E[j].u;int v=E[j].v;double a=E[j].a;double b=E[j].b;if(dist[v]<(dist[u]-b)*a){dist[v]=(dist[u]-b)*a;flag=true;}}if(!flag)return true;//没有负环回路}for(int j=0; j<E.size(); j++)if(dist[E[j].v]<(dist[E[j].u]-E[j].b)*E[j].a)return false;//有负环回路return true;//没有负环回路
}
int main()
{
#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);
#endifint n,m,s;double x;while(scanf("%d%d%d%lf",&n,&m,&s,&x)!=EOF){int a,b;double w1,w2,w3,w4;E.clear();for(int i=0; i<m; i++){scanf("%d%d%lf%lf%lf%lf",&a,&b,&w1,&w2,&w3,&w4);E.push_back(Edge(a,b,w1,w2));E.push_back(Edge(b,a,w3,w4));}if(bellman_ford(s,n,x))printf("NO\n");else puts("YES");}return 0;
}