poj1860 & poj2240(Bellman-Ford)

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;
}