POJ1860《Currency Exchange》方法：反向Bellman-ford

就是求一条正权回环，方法在于反向使用bellman-ford。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int n; //货币种数int m; //交换点int s; //目前持有的货币种类double v; //持有货币钱数int all=0; //边数double dis[101]; //源点到各点的权值class exchange_point{public:int a;int b;double r;double c;} exc[201];bool bellman_ford(){memset(dis, 0, sizeof(dis));dis[s] = v; // 初始源点for (int i = 1; i <= n-1; ++i) { // |V|-1次，点数bool flag = false; // 用于截断无法正向松弛，避免无谓的操作for (int j = 0; j < all; ++j) { // 对于每一条边进行|V|次松弛,因为是求正权回环if (dis[exc[j].b] < (dis[exc[j].a] - exc[j].c) * exc[j].r) {dis[exc[j].b] = (dis[exc[j].a] - exc[j].c) * exc[j].r;flag = true;}}if (!flag)break;}//if (dis[s] > v) // 这样不行//return true;for (int i = 0; i < all; ++i) //找到正环能无限松弛，找到正权回路if (dis[exc[i].b] < (dis[exc[i].a] - exc[i].c) * exc[i].r)return true;return false;}int main(){//freopen("temp.txt", "r", stdin);int a, b;double rab, cab, rba, cba; // ratio, commissioncin >> n >> m >> s >> v;for (int i = 0; i < m; ++i) {cin >> a >> b >> rab >> cab >> rba >> cba;exc[all].a = a;exc[all].b = b;exc[all].r = rab;exc[all++].c = cab;exc[all].a = b;exc[all].b = a;exc[all].r = rba;exc[all++].c = cba;}if (bellman_ford())cout << "YES" << endl;else cout << "NO" << endl;return 0;}