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POJ 2240 Arbitrage
Arbitrage
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 16175 | Accepted: 6803 |
Description
Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British pound buys 10.0 French francs, and 1 French franc buys 0.21 US dollar. Then, by converting currencies, a clever trader can start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5 percent.
Your job is to write a program that takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.
Your job is to write a program that takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.
Input
The input will contain one or more test cases. Om the first line of each test case there is an integer n (1<=n<=30), representing the number of different currencies. The next n lines each contain the name of one currency. Within a name no spaces will appear. The next line contains one integer m, representing the length of the table to follow. The last m lines each contain the name ci of a source currency, a real number rij which represents the exchange rate from ci to cj and a name cj of the destination currency. Exchanges which do not appear in the table are impossible.
Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n.
Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n.
Output
For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes" respectively "Case case: No".
Sample Input
3USDollarBritishPoundFrenchFranc3USDollar 0.5 BritishPoundBritishPound 10.0 FrenchFrancFrenchFranc 0.21 USDollar3USDollarBritishPoundFrenchFranc6USDollar 0.5 BritishPoundUSDollar 4.9 FrenchFrancBritishPound 10.0 FrenchFrancBritishPound 1.99 USDollarFrenchFranc 0.09 BritishPoundFrenchFranc 0.19 USDollar0
Sample Output
Case 1: YesCase 2: No
Source
Ulm Local 1996
将字符串用map处理一下。
然后用spfa判一下环就可以了~
#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <vector>#include <queue>#include <map>#include <set>#include <stack>#include <algorithm>using namespace std;typedef long long LL;const int N = 110;const int inf = 1e7+7;const double PI = acos(-1.0);const double eps = 1e-6 ;int n , m ;struct node { int v ; double w ; node(){} node( int a , double b ) { v = a , w = b ; }};vector<node>g[N];map<string,int>mp;string str1 , str2 ;bool vis[N] , inq[N] ;double dis[N];int in_cnt[N];bool spfa( int s ) { queue<int>que; for( int i = 0 ; i <= n ; ++i ) dis[i] = 0.0 , in_cnt[i] = 0 , inq[i] = false; dis[s] = 1.0 ; inq[s] = true ; que.push(s); while( !que.empty() ) { int u = que.front() ; que.pop(); inq[u] = false , vis[u] = true ; for( int i = 0 ; i < g[u].size() ; ++i ) { int v = g[u][i].v ; double w = g[u][i].w ; if( dis[u] * w > dis[v] ) { dis[v] = dis[u] * w ; if( !inq[v] ) { in_cnt[v] ++ ; if( in_cnt[v] >= n ) return false; inq[v] = true; que.push(v); } } } } return true;}void run(){ double w ; mp.clear(); for( int i = 0 ; i <= n ; ++i )g[i].clear(); for( int i = 1 ; i <= n ; ++i ){ cin >> str1 , mp[str1] = i ; } cin >> m ; while( m-- ) { cin >> str1 >> w >> str2 ; int u = mp[str1] , v = mp[str2] ; g[u].push_back( node(v,w) ); } memset(vis,false,sizeof vis); bool tag = false ; for( int i = 1 ; i <= n ; ++i ) if( !vis[i] ) { if( !spfa(i) ) { tag = true ; break ; } } if( tag ) cout << "Yes" <<endl; else cout << "No" << endl;}int main(){ #ifdef LOCAL freopen("in.txt","r",stdin); #endif // LOCAL ios::sync_with_stdio(false); int cas = 1 ; while( cin >> n && n ){ cout << "Case "<<cas++ <<": ";run(); }}
POJ 2240 Arbitrage
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