HDU 1271 Arbitrage

链接:

http://acm.hdu.edu.cn/showproblem.php?pid=1217

题目：

Arbitrage

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2594    Accepted Submission(s): 1167

Problem 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 file 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

University of Ulm Local Contest 1996

 

Recommend

Eddy

 

分析与总结：

根据题意可知，只要找到正环便可判断为Yes，因此可以SPFA判环，或者建好邻接矩阵图用Floyd算法计算后，判断是否有d[i][i]>1即可。

1.Floyd

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<map>
#include<string>
using namespace std;

typedef double Type;
const int INF = 0x7ffffff;
const int VN  = 35;
const int EN  = VN*VN;

map<string, int>mp;
int n,size,cnt;
int head[VN];
Type d[VN][VN];

void init(){
    size=0; cnt=0;
    mp.clear();
    for(int i=1; i<=n; ++i){
        d[i][i] = 1;
        for(int j=i+1; j<=n; ++j)
            d[i][j] = d[j][i] = 0;
    }
}

void Floyd(){
    for(int k=1; k<=n; ++k)
    for(int i=1; i<=n; ++i)
    for(int j=1; j<=n; ++j){
        d[i][j] = max(d[i][j], d[i][k]*d[k][j]);
    }
}


int main(){
    char str1[50], str2[50];
    int cas=1,m,u,v;
    double w;

    while(scanf("%d",&n)&&n){
        init();
        for(int i=0; i<n; ++i){
            scanf("%s",str1);
            mp[str1] = ++cnt;
        }
        scanf("%d",&m);
        for(int i=0; i<m; ++i){
            scanf("%s %lf %s",str1,&w,str2);
            u = mp[str1], v = mp[str2];
            d[u][v] = w;
        }
        bool flag=false;
        Floyd();
        for(int i=1; i<=n; ++i)
            if(d[i][i]>1.0) {flag=true; break;}

        printf("Case %d: ",cas++);
        if(flag) puts("Yes");
        else puts("No");
    }
    return 0;
}

2.SPFA判正环

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<queue>
#include<map>
using namespace std;

typedef double Type;
const int INF = 0x7ffffff;
const int VN  = 35;
const int EN  = VN*VN;

map<string, int>mp;
int n,size,cnt;
int head[VN];
bool inq[VN];
int counter[VN];
Type d[VN];

struct Edge{int v,next; Type w;}E[EN];
void addEdge(int u,int v,Type w){
    E[size].v=v, E[size].w=w;
    E[size].next = head[u];
    head[u] = size++;
}
void init(){
    size=0; cnt=0;
    memset(head, -1, sizeof(head));
    mp.clear();
}

bool SPFA(int src){
    memset(counter, 0, sizeof(counter));
    memset(inq, 0, sizeof(inq));
    for(int i=1; i<=n; ++i) d[i]=0;
    d[src] = 1;
    queue<int>q;
    q.push(src);
    while(!q.empty()){
        int u = q.front();  q.pop();
        inq[u] = false;
        for(int e=head[u]; e!=-1; e=E[e].next){
            Type tmp = d[u]*E[e].w;
           // if(d[E[e].v] > tmp){
            if(d[E[e].v]<tmp){
                d[E[e].v] = tmp;
                if(!inq[E[e].v]){
                    inq[E[e].v] = true;
                    q.push(E[e].v);
                    if(++counter[E[e].v]>=n){
                        return true;
                    }
                }
            }
            
        }
    }
    return false;
}


int main(){
    char str1[50], str2[50];
    int cas=1,m,u,v;
    double w;

    while(scanf("%d",&n)&&n){
        init();
        for(int i=0; i<n; ++i){
            scanf("%s",str1);
            mp[str1] = ++cnt;
        }
        scanf("%d",&m);
        for(int i=0; i<m; ++i){
            scanf("%s %lf %s",str1,&w,str2);
            u = mp[str1], v = mp[str2];
            addEdge(u,v,w);
        }
        bool flag=false;
        for(int i=1; i<=n; ++i){
            if(SPFA(i)){
                flag=true;
                break;
            }
        }
        printf("Case %d: ",cas++);
        if(flag) puts("Yes");
        else puts("No");
    }
    return 0;
}

  
   
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        原创 http://blog.csdn.net/shuangde800 ， By   D_Double  (转载请标明)