Currency Exchange (POJ 1860)

Currency Exchange

思路：用spfa来找正环，只要存在一个正环（cnt >= n），那么走这个圈无穷次后资产一定正无穷，因为路是双向的，所以一定可以返回原来的货币，正无穷再怎么亏损,一定还是正无穷

AC Code

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;const int N = 2021;
int e[N], ne[N], h[N], idx;
double rate[N], cost[N]; //汇率 佣金
double money[N];
int n, m, s; //货币种类， 兑换地点， 当前钱种类
double v;   // 当前钱数量
bool st[N];
int cnt[N];void add(int a, int b, double r, double c)
{e[idx] = b, rate[idx] = r, cost[idx] = c, ne[idx] = h[a]; h[a] = idx++;
}bool spfa()
{queue<int> q;q.push(s);st[s] = true;cnt[s] = 0;money[s] = v;while(!q.empty()){int t = q.front();q.pop();st[t] = false;for(int i = h[t]; i != -1; i = ne[i]){int j = e[i];double tmp = (money[t] - cost[i]) * rate[i];if(tmp > money[j]){money[j] = tmp;cnt[j] = cnt[t] + 1;if(cnt[j] >= n)return true;if(!st[j]){q.push(j);st[j] = true;}}}} return false;
}int main()
{cin >> n >> m >> s >> v;memset(h, -1, sizeof h);while(m--){int a, b;double r1, c1, r2, c2;cin >> a >> b >> r1 >> c1 >> r2 >> c2;add(a, b, r1, c1);add(b, a, r2, c2);}cout << (spfa() ? "YES" : "NO") << endl;return 0;
}

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