http://poj.org/problem?id=2240
题意：
货币兑换，判断最否是否能获利。
思路：
又是货币兑换题，Belloman-ford和floyd算法都可以的。
1 #include<iostream> 2 #include<algorithm> 3 #include<string> 4 #include<cstring> 5 #include<map> 6 using namespace std; 7 8 const int maxn = 1000 + 5; 9 10 int n, m; 11 string s1,s2; 12 13 map<string, int> p; 14 15 struct 16 { 17 int u, v; 18 double rate; 19 }edge[maxn]; 20 21 double d[40]; 22 23 bool bellman(int s) 24 { 25 memset(d, 0, sizeof(d)); 26 d[s] = 1; 27 for (int i = 1; i < n; i++) 28 { 29 bool flag = false; 30 for (int j = 0; j < m; j++) 31 { 32 if (d[edge[j].v] < d[edge[j].u] * edge[j].rate) 33 { 34 d[edge[j].v] = d[edge[j].u] * edge[j].rate; 35 flag = true; 36 } 37 } 38 if (!flag) break; 39 } 40 for (int j = 0; j < m; j++) 41 if(d[edge[j].v] < d[edge[j].u] * edge[j].rate) 42 return true; 43 return false; 44 } 45 46 int main() 47 { 48 ios::sync_with_stdio(false); 49 //freopen("D:\txt.txt", "r", stdin); 50 double r; 51 int kase = 0; 52 while (~scanf("%d", &n) && n) 53 { 54 for (int i = 0; i < n; i++) 55 { 56 cin >> s1; 57 p[s1] = i; 58 } 59 scanf("%d", &m); 60 for (int i = 0; i < m; i++) 61 { 62 cin >> s1 >> r >> s2; 63 edge[i].u = p[s1]; 64 edge[i].v = p[s2]; 65 edge[i].rate = r; 66 } 67 printf("Case %d: ", ++kase); 68 bool flag = false; 69 for (int i = 0; i < n; i++) 70 { 71 if (bellman(i)) 72 { 73 printf("Yes "); 74 flag = true; 75 break; 76 } 77 } 78 if (!flag) 79 printf("No "); 80 } 81 return 0; 82 }
1 #include<iostream> 2 #include<algorithm> 3 #include<string> 4 #include<cstring> 5 #include<map> 6 using namespace std; 7 8 const int maxn = 1000 + 5; 9 const int INF = 10000; 10 11 int n, m; 12 string s1,s2; 13 14 map<string, int> p; 15 16 double d[maxn][maxn]; 17 18 void floyd() 19 { 20 for (int k = 0; k < n; k++) 21 for (int i = 0; i < n; i++) 22 for (int j = 0; j < n; j++) 23 d[i][j] = max(d[i][j], d[i][k] * d[k][j]); 24 } 25 26 int main() 27 { 28 ios::sync_with_stdio(false); 29 //freopen("D:\txt.txt", "r", stdin); 30 double r; 31 int kase = 0; 32 while (~scanf("%d", &n) && n) 33 { 34 memset(d, INF, sizeof(d)); 35 for (int i = 0; i < n; i++) 36 { 37 cin >> s1; 38 p[s1] = i; 39 d[i][i] = 1; //一开始的汇率为1 40 } 41 scanf("%d", &m); 42 for (int i = 0; i < m; i++) 43 { 44 cin >> s1 >> r >> s2; 45 d[p[s1]][p[s2]] = r; 46 } 47 floyd(); 48 printf("Case %d: ", ++kase); 49 bool flag = false; 50 for (int i = 0; i < n; i++) 51 { 52 if (d[i][i]>1) //如果自身汇率大于1，说明可以套利 53 { 54 printf("Yes "); 55 flag = true; 56 break; 57 } 58 } 59 if (!flag) 60 printf("No "); 61 } 62 return 0; 63 }