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;
}