POJ 2240 Arbitrage

[★★☆☆☆]图论 最短路

• 题目大意：

  可以简单描述为知道从i到j的汇率，问能不能赚钱

• 样例

  输入：
  USDollar
  BritishPound
  FrenchFranc
  3
  USDollar 0.5 BritishPound
  BritishPound 10.0 FrenchFranc
  FrenchFranc 0.21 USDollar

  3
  USDollar
  BritishPound
  FrenchFranc
  6
  USDollar 0.5 BritishPound
  USDollar 4.9 FrenchFranc
  BritishPound 10.0 FrenchFranc
  BritishPound 1.99 USDollar
  FrenchFranc 0.09 BritishPound
  FrenchFranc 0.19 USDollar

  0

  输出：
  Case 1: Yes
  Case 2: No

• 解题思路：

  水题，本来能1A的，结果交上去超时。
  看了Discuss是编译器的问题，我本来是G++换成C++就A了。
  我用的Bellman算法，Floyd也行。

• 代码

#include <iostream>#include <algorithm>#include <string>using namespace std;struct edge {    int from, to;    double r;};edge E[1000];int cte;bool used[35];string S[35];double d[35];int V;int nS(string s) {    for (int i = 1; i <= V; i++) {        if (s == S[i]) return i;    }    return 0;}void adde(int f, int t, double r) {    edge te = {f, t, r};    E[cte++] = te;}int usedall() {    for (int i = 1; i <= V; i++) {        if (d[i] == 0) return i;    }    return 0;}bool bellman() {    int u;    for (int i = 1; i <= V; i++) d[i] = 0;    while ((u = usedall())) {        d[u] = 1;        int ct;        for (ct = 1; ct <= V; ct++) {            bool update = false;            for (int i = 0; i < cte; i++) {                edge e = E[i];                if (d[e.from] != 0 && d[e.to] < d[e.from] * e.r) {                    d[e.to] = d[e.from] * e.r;                    update = true;                }            }            if (!update) break;        }        if (ct == V+1) return 1;    }    return 0;}int main() {    int cs = 1;    while ((cin >> V) && V != 0) {        cte = 0;        for (int i = 1; i <= V; i++) {            cin >> S[i];        }        int n;        string s1, s2;        double r;        cin >> n;        for (int i = 1; i <= n; i++) {            cin >> s1 >> r >> s2;            adde(nS(s1), nS(s2), r);        }        cout << "Case " << cs++;        if (bellman()) cout << ": Yes" << endl;        else cout << ": No" << endl;    }    return 0;}