题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1217
题意：给一些汇率，问是否可以套利。
套利就是经过一些汇率的转换，最终使得本金额比起始金额大。
最短路跑所有的汇率情况，看看有没有使得最终的汇率是大于1的。
1 /* 2 ━━━━━┒ギリギリ♂ eye！ 3 ┓┏┓┏┓┃キリキリ♂ mind！ 4 ┛┗┛┗┛┃＼○／ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <cassert> 24 #include <cstdio> 25 #include <bitset> 26 #include <vector> 27 #include <deque> 28 #include <queue> 29 #include <stack> 30 #include <ctime> 31 #include <set> 32 #include <map> 33 #include <cmath> 34 using namespace std; 35 #define fr first 36 #define sc second 37 #define cl clear 38 #define BUG puts("here!!!") 39 #define W(a) while(a--) 40 #define pb(a) push_back(a) 41 #define Rint(a) scanf("%d", &a) 42 #define Rs(a) scanf("%s", a) 43 #define FRead() freopen("in", "r", stdin) 44 #define FWrite() freopen("out", "w", stdout) 45 #define Rep(i, len) for(int i = 0; i < (len); i++) 46 #define For(i, a, len) for(int i = (a); i < (len); i++) 47 #define Cls(a) memset((a), 0, sizeof(a)) 48 #define Clr(a, x) memset((a), (x), sizeof(a)) 49 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 50 #define lrt rt << 1 51 #define rrt rt << 1 | 1 52 #define pi 3.14159265359 53 #define RT return 54 #define lowbit(x) x & (-x) 55 #define onenum(x) __builtin_popcount(x) 56 typedef long long LL; 57 typedef long double LD; 58 typedef unsigned long long ULL; 59 typedef pair<int, int> pii; 60 typedef pair<string, int> psi; 61 typedef pair<LL, LL> pll; 62 typedef map<string, int> msi; 63 typedef vector<int> vi; 64 typedef vector<LL> vl; 65 typedef vector<vl> vvl; 66 typedef vector<bool> vb; 67 68 const int maxn = 33; 69 const double eps = 1e-9; 70 map<string, int> name; 71 double dp[maxn][maxn]; 72 int n, m; 73 char a[81], b[81]; 74 double ex; 75 76 int main() { 77 // FRead(); 78 int _ = 1; 79 while(~Rint(n) && n) { 80 int cnt = 1; 81 name.clear(); Cls(dp); 82 For(i, 1, n+1) { 83 Rs(a); 84 name[a] = i; 85 dp[i][i] = 1.0; 86 } 87 Rint(m); 88 Rep(i, m) { 89 cin >> a >> ex >> b; 90 dp[name[a]][name[b]] = ex; 91 } 92 For(k, 1, n+1) { 93 For(i, 1, n+1) { 94 For(j, 1, n+1) { 95 if(dp[i][k] * dp[k][j] > dp[i][j]) { 96 dp[i][j] = dp[i][k] * dp[k][j]; 97 } 98 } 99 } 100 } 101 bool flag = 0; 102 For(i, 1, n+1) { 103 if(dp[i][i] > 1) { 104 flag = 1; 105 printf("Case %d: Yes ", _++); 106 break; 107 } 108 } 109 if(!flag) printf("Case %d: No ", _++); 110 } 111 RT 0; 112 }