1860的思路是将可以换得的不同种的货币的数量当作节点,每个兑换点当成边,然后我抄了个算法导论里面的Bellman-Ford算法,一次就过了。看discussion里面很多讨论精度的,我想都没想过……
2240是更简单的一个bellman-ford,基本和1860一样,就只多了个map容器的应用而已。
以下是1860:
#include <iostream> using namespace std; struct Point{ int a, b; double Rab, Cab, Rba, Cba; }; int main() { int n, m, s; double money; Point point[101]; cin >> n >> m >> s >> money; for (int i = 0; i < m; i++){ cin >> point[i].a >> point[i].b >> point[i].Rab >> point[i].Cab >> point[i].Rba >> point[i].Cba; } double node[101]; memset(node, 0, sizeof(node)); node[s] = money; for (int j = 1; j <= n - 1; j++){ for (int i = 0; i < m; i++){ if (node[point[i].a] < (node[point[i].b] - point[i].Cba) * point[i].Rba) node[point[i].a] = (node[point[i].b] - point[i].Cba) * point[i].Rba; if (node[point[i].b] < (node[point[i].a] - point[i].Cab) * point[i].Rab) node[point[i].b] = (node[point[i].a] - point[i].Cab) * point[i].Rab; } } bool flag = true; for (int i = 0; i < m; i++){ if (node[point[i].a] < (node[point[i].b] - point[i].Cba) * point[i].Rba || node[point[i].b] < (node[point[i].a] - point[i].Cab) * point[i].Rab){ flag = false; break; } } cout << (flag ? "NO" : "YES") << endl; return 0; }
2240:
#include <iostream> #include <map> #include <string> using namespace std; struct Edge{ int type1, type2; double rate; }; int main() { int n; int testCase = 1; while (cin >> n && n != 0){ double node[31]; Edge edge[1000]; map<string, int> currency; for (int i = 0; i < n; i++){ string type; cin >> type; currency[type] = i; } int m; cin >> m; for (int i = 0; i < m; i++){ string type1, type2; double r; cin >> type1 >> r >> type2; edge[i].type1 = currency[type1]; edge[i].type2 = currency[type2]; edge[i].rate = r; } for (int i = 0; i < n; i++){ node[i] = 1000000; } node[0] = 1; for (int i = 1; i <= n - 1; i++){ for (int j = 0; j < m; j++){ double tmp = node[edge[j].type1] * edge[j].rate; if (node[edge[j].type2] < tmp){ node[edge[j].type2] = tmp; } } } bool flag = false; for (int i = 0; i < m; i++){ if (node[edge[i].type2] < node[edge[i].type1] * edge[i].rate){ flag = true; break; } } cout << "Case " << testCase << ": " << (flag ? "Yes" : "No") << endl; testCase++; } return 0; }