POJ1860 Currency Exchange(判断正环)

题意：

n种货币，m个兑换点，每中方兑换都有汇率和手续费，初始有s类型的货币v元，兑换后钱数为（当前钱 - 手续费）*汇率。问是否可以让钱增加。

思路：

SPFA判断是否存在正环

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int maxn = 100+10;
using namespace std;

int n,m;
double dis[maxn];
// 图
int head[maxn], cur; 
struct Edge{
	int v, next;
	double rate, cost;
};
Edge edges[10000];
void init(int a){
	for(int i = 0; i <= a; ++i) head[i] = -1;
	cur = 0;
}
void addEdge(int u, int v, double r1, double c1, double r2, double c2){
	edges[cur].v = v; edges[cur].rate = r1; edges[cur].cost = c1;
	edges[cur].next = head[u]; head[u] = cur++;
	edges[cur].v = u; edges[cur].rate = r2; edges[cur].cost = c2;
	edges[cur].next = head[v]; head[v] = cur++;
}


bool inq[maxn];
int cnt[maxn];

bool BellmanFord(int s){
	memset(inq, 0, sizeof(inq)); inq[s] = 1;
	memset(cnt, 0, sizeof(cnt)); cnt[s] = 1;
	queue<int> Q;
	Q.push(s);
	while(!Q.empty()){
		int x = Q.front(); Q.pop();
		inq[x] = 0;
		for(int i = head[x]; i != -1; i = edges[i].next){
			Edge& e = edges[i];
			if(dis[e.v] < (dis[x] - e.cost)*e.rate){
				dis[e.v] = (dis[x] - e.cost)*e.rate;
				if(!inq[e.v]){ Q.push(e.v) ; inq[e.v] = 1; if(++cnt[e.v] > n) return 1; } 
			}
		}
	}
	return 0;
}

int main()
{
    //freopen("in.txt","r",stdin);
    int s, a, b;
    double v,c,r,e,f;
	
	scanf("%d %d %d %lf",&n,&m,&s,&v);
	init(n);
	memset(dis, 0, sizeof(dis));
	dis[s] = v;
	for(int i = 0; i < m; ++i){
		scanf("%d %d %lf %lf %lf %lf", &a, &b, &c, &r, &e, &f);
		addEdge(a,b,c,r,e,f);
	}
	
	if(BellmanFord(s)) printf("YES\n");
	else printf("NO\n");
	
    //fclose(stdin);
	return 0;
}