POJ 2240 Arbitrage 最短路径 Floyed-Warshall

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18416		Accepted: 7800

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.  

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.

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

3
USDollar
BritishPound
FrenchFranc
3
USDollar 0.5 BritishPound
BritishPound 10.0 FrenchFranc
FrenchFranc 0.21 USDollar

3
USDollar
BritishPound
FrenchFranc
6
USDollar 0.5 BritishPound
USDollar 4.9 FrenchFranc
BritishPound 10.0 FrenchFranc
BritishPound 1.99 USDollar
FrenchFranc 0.09 BritishPound
FrenchFranc 0.19 USDollar

0

Sample Output

Case 1: Yes
Case 2: No

Source

Ulm Local 1996

题目大意：

不同货币间有不同的汇率，问是否存在货币a-->货币b--->.......----> 货币a实现获利

可以转化为加权有向图货币对应节点，货币兑换关系对应边

则货币i经由货币k兑换至货币j的比率为 dist[i][k]*dist[k][j]

\因为本题含回路所以要加条件    i!=j&&k!=j&&k!=i  防止死循环

最后判断 dist[i][j]*dist[j][i]>1 表示可以套汇

ACcode:

#pragma warning(disable:4786)//使命名长度不受限制
#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈
#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rds(x) scanf("%s",x)
#define rdc(x) scanf("%c",&x)
#define ll long long int
#define maxn 1005
#define mod 1000000007
#define INF 0x3f3f3f3f //int 最大值
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define MT(x,i) memset(x,i,sizeof(x))
#define PI  acos(-1.0)
#define E  exp(1)
#define esp 1e-6
using namespace std;
char money[maxn][55],x[55],y[55];
double dist[maxn][maxn],t;
int n,m;
int get_id(char *str){
    FOR(i,1,n)if(!strcmp(money[i],str))return i;
}
int main(){
    int cnt=1;
    while(rd(n)&&n){
        MT(dist,0);
        FOR(i,1,n)rds(money[i]);
        rd(m);FOR(i,1,m){
            scanf("%s%lf%s",&x,&t,&y);
            dist[get_id(x)][get_id(y)]=t;
            }
        FOR(k,1,n)FOR(i,1,n)FOR(j,1,n)
            if(i!=j&&k!=j&&k!=i)
                dist[i][j]=max(dist[i][k]*dist[k][j],dist[i][j]);
        bool flag=false;
        FOR(i,1,n)FOR(j,1,n)
            if(dist[i][j]*dist[j][i]>1+esp){
                flag=true;break;
            }
        printf("Case %d: %s",cnt++,flag?"Yes\n":"No\n");
    }
    return 0;
}
/*
3
USDollar
BritishPound
FrenchFranc
3
USDollar 0.5 BritishPound
BritishPound 10.0 FrenchFranc
FrenchFranc 0.21 USDollar

3
USDollar
BritishPound
FrenchFranc
6
USDollar 0.5 BritishPound
USDollar 4.9 FrenchFranc
BritishPound 10.0 FrenchFranc
BritishPound 1.99 USDollar
FrenchFranc 0.09 BritishPound
FrenchFranc 0.19 USDollar

0
*/