I - Arbitrage
题目链接：https://vjudge.net/contest/66569#problem/I
题目：
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 题意：给你n种货币，和m种货币转换关系还有之间的单位货币数，问能否经过这些关系使得原先货数增加思路：题没有给原先货币数，故令原先货币数为1，这题不是计算路权和而是路权积，利用spfa算法即可快速判断，由于数据范围小可直接用邻接矩阵来存图，用时79s
//
// Created by hy on 2019/7/20.
//
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <queue>
#include <map>
#include<memory.h>
using namespace std;
const int maxn=105;
#define MAX 0x3f3f3f3f
int n,m;
double road[maxn][maxn];
double d[maxn];
int book[maxn];
int rnt[maxn];
bool spfa(int s)
{
memset(book,0,sizeof(book));
memset(d,MAX,sizeof(d));
memset(rnt,0,sizeof(rnt));
queue<int>qu;
qu.push(s);
book[s]=1;
d[s]=1;
rnt[s]++;
while(!qu.empty())
{
int now=qu.front();
qu.pop();
book[now]=0;
for(int i=1;i<=n;i++)
{
if(d[i]<d[now]*road[now][i])
{
d[i]=d[now]*road[now][i];
if(!book[i])
{
qu.push(i);
book[i]=1;
rnt[i]++;
if(rnt[i]>n-1)
return true;
}
}
}
}
return false;
}
int main() {
int res=0;
while (~scanf("%d", &n)) {
if(n==0)
break;
string str;
map<string, int> mp;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=i;j++)
{
if(i==j)
road[i][j]=0;
else
road[i][j]=MAX;
}
}
for (int i = 1; i <= n; i++) {
cin >> str;
mp[str] = i;
}
string str1, str2;
double num;
scanf("%d",&m);
for (int i = 1; i <= m; i++) {
cin >> str1 >> num >> str2;
road[mp[str1]][mp[str2]] = num;
}
res++;
if (spfa(1))
printf("Case %d: Yes\n",res);
else
printf("Case %d: No\n",res);
}
return 0;
}
原文地址：https://www.cnblogs.com/Vampire6/p/11219628.html