Arbitrage
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 29374 | Accepted: 12279 |
题目链接：http://poj.org/problem?id=2240
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
题意：
给出几种货币单位，以及货币与货币之间的兑换汇率，问最后是否能够套利。就是用1个单位的货币，不断去兑换其它的货币，最后得到大于1个单位的相应货币。
题解：
总的思路就是跑最长路看看是否有正环吧，有的话就说明至少存在一种货币可以用来套利。
这里跑最长路的时候要把之前的“ + ”改造为“ * ”，至于正确性，乘法取个对数也等价于加吧？具体证明我也不是很清楚。
代码如下：
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <string>
#include <stack>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 35;
map <string ,int> mp;
int n,num;
struct Edge{
int u,v,next;
double w;
}e[N*N<<1];
int head[N],vis[N],c[N];
int tot;
void adde(int u,int v,double w){
e[tot].v=v;e[tot].w=w;e[tot].next=head[u];head[u]=tot++;
}
double d[N];
int spfa(int s){
memset(d,0,sizeof(d));memset(vis,0,sizeof(vis));
d[s]=1;queue <int> q;q.push(s);memset(c,0,sizeof(c));
c[s]=1;vis[s]=1;
while(!q.empty()){
int u=q.front();q.pop();vis[u]=0;
for(int i=head[u];i!=-1;i=e[i].next){
int v=e[i].v;
if(d[v]<d[u]*e[i].w){
d[v]=d[u]*e[i].w;
if(!vis[v]){
vis[v]=1;
q.push(v);
if(++c[v]>n) return 1;
}
}
}
}
return 0;
}
int main(){
int cnt =0;
while(scanf("%d",&n)!=EOF){
if(n==0) break ;
string s;
num=0;cnt++;
for(int i=1;i<=n;i++){
cin>>s;
mp[s]=++num;
}
int tmp;
scanf("%d",&tmp);
memset(head,-1,sizeof(head));tot=0;
for(int i=1;i<=tmp;i++){
string s1,s2;
double w;
cin>>s1>>w>>s2;
adde(mp[s1],mp[s2],w);
}
printf("Case %d: ",cnt);
if(spfa(1)) puts("Yes");
else puts("No");
}
return 0;
}
原文地址：https://www.cnblogs.com/heyuhhh/p/10352067.html