Arbitrage poj(贝尔曼福特判正权环+串)

Arbitrage

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17691		Accepted: 7467

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

套汇问题。。。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
struct node
{
    char  s[16];
    int num;
} qu[1000];
struct no//注意q[]的数组应该开大点，最环的情况为n*n
{
    int a,b;
    double r;
} q[1000];

char s1[16],s2[16];
double rate;
int   sum,n,m;
double  dis[1000000];

bool  bf()
{
    int i,j;
    memset(dis,0,sizeof(dis));
    dis[1]= 1  ;//一般题目没有名白的说从哪里开始，墨认为1
    for(i=1; i<n; i++)
        for(j=0; j<sum; j++)
            if(dis[q[j].b  ]<dis[ q[j].a  ]*q[j].r )
                dis[q[j].b   ]=dis[  q[j].a ]*q[j].r;
        for(j=0; j<sum; j++)//一定注意再判断正环时，只在q[]自己的范围内，而不是
            if(dis[q[j].b  ]<dis[ q[j].a  ]*q[j].r)
            {
                return   true ;
            }
    return false;
}
int cla=1;
int main()
{
    ios::sync_with_stdio(false);
    while(scanf("%d",&n)&&n)
    {
        sum=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%s",qu[i].s);
            qu[i].num=i;
        }
        scanf("%d",&m);
        while(m--)
        {
            int a=0,b=0;
            scanf("%s%lf%s",s1,&rate,s2);
            for(int j=1; j<=n; j++)
            {
                if( ! strcmp(s1,qu[j].s)   )
                    a=qu[j].num;
                if(  !strcmp(s2,qu[j].s)   )
                    b=qu[j].num;
                if(a!=0&&b!=0)
                    break;
            }
            q[sum].a=a;
            q[sum].b=b;
            q[sum++].r=rate;
        }
        bool  o=bf();
        printf("Case %d: ",cla++);
        if(o==true)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}

SPFA：

#include<map>
#include<queue>
#include<cmath>
#include<iostream>
#include<cstdio>
#include<stack>
#include<cstring>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
char s[1000],s1[1000];
double Map[100][100],dis[100];

int m,n,tmp[100],sum;
bool vis[100];
bool spfa(){
    for(int i=0;i<=n;i++){
        dis[i]=0;
        vis[i]=false;
        tmp[i]=0;
    }

    dis[1]=1;
    vis[1]=true;
    queue<int>q;while(!q.empty())q.pop();
    q.push(1);
    while(!q.empty()){
        int u=q.front();q.pop();
        tmp[u]++;
        if(tmp[u]>n) return false;
        vis[u]=false;
        for(int i=1;i<=n;i++){
            if( (dis[i]<dis[u]*Map[u][i])&&Map[u][i]){
                dis[i]=dis[u]*Map[u][i];
                if(!vis[i]){
                    vis[i]=true;
                    q.push(i);
                }
            }
        }
    }
    return true;
}
int main(){
    int i,j,k,cnt=1;
    while(~scanf("%d",&n)&&n){
        sum=0;
        map<string,int>mp;
        mp.clear();
        for(i=1;i<=n;i++){
            scanf("%s",s);
            mp[s]=i;
        }
        for(i=0;i<100;i++){
            for(j=0;j<100;j++){
                Map[i][j]=0;
            }
        }
        scanf("%d",&m);
        double p;
        for(i=1;i<=m;i++){
            scanf("%s %lf %s",s,&p,s1);
            Map[mp[s]][mp[s1]]=p;
        }
        bool o=spfa();
        printf("Case %d: ",cnt++);
        if(!o){
            puts("Yes");
        }
        else
            puts("No");
    }
    return 0;
}