poj_2240 Arbitrage(Bellman_ford + map)
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Arbitrage
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 21409 Accepted: 9122
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
用上map简单地处理字符串,然后就是反向ballmen-ford找正环了。
用上cin会慢上几个倍数,还是尽量用scanf了
#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <stack>#include <bitset>#include <queue>#include <set>#include <map>#include <string>#include <algorithm>#define FOP freopen("data.txt","r",stdin)#define inf 0x3f3f3f3f#define maxn 32#define mod 1000000007#define PI acos(-1.0)#define LL long longusing namespace std;struct Edge{ int a, b; double c;}edge[maxn * maxn];double dis[maxn];int n, m;map<string, int>M;bool bellman_ford(int s){ for(int i = 1; i <= n; i++) dis[i] = 0; dis[s] = 1; for(int i = 1; i <= n - 1; i++) { for(int j = 1; j <= m; j++) { int a = edge[j].a, b = edge[j].b; double t = dis[a] * edge[j].c; if(dis[b] < t) dis[b] = t; } } for(int i = 1; i <= m; i++) { int a = edge[i].a, b = edge[i].b; double t = dis[a] * edge[i].c; if(dis[b] < t) return true; } return false;}int main(){ int casen = 0; string s, s2; while(scanf("%d", &n) && n) { casen++; for(int i = 1; i <= n; i++) { //cin>>s; char ss[100]; scanf("%s", ss); s = ss; M[s] = i; } scanf("%d", &m); int a, b; double c; for(int i = 1; i <= m; i++) { //cin>>s>>c>>s2; char ss1[100], ss2[100]; scanf("%s%lf%s", ss1, &c, ss2); s = ss1; s2 = ss2; a = M[s], b = M[s2]; edge[i].a = a, edge[i].b = b, edge[i].c = c; } bool flag = false; for(int i = 1; i <= n; i++) { if(bellman_ford(i)) { printf("Case %d: Yes\n", casen); flag = true; break; } } if(!flag) printf("Case %d: No\n", casen); } return 0;}
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