POJ-2240-Arbitrage（弗洛伊德|贝尔曼）

Arbitrage
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 20786 Accepted: 8859
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

坑点：建图时可能出现自己到自己的情况

代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
//先用弗洛伊德
const int maxn=33;
char str[maxn][maxn];//字典
double map[maxn][maxn];
int main()
{
    int N;
    int casen=1;
    while(~scanf("%d",&N)&&N)
    {
        for(int i=1; i<=N; i++)
            scanf("%s",str[i]);
        int M;
        scanf("%d",&M);
        char str_1[maxn];
        double num;
        char str_2[maxn];
        for(int i=0; i<=N; i++)//初始化图
            for(int j=0; j<=N; j++)
                map[i][j]=0.0;
        for(int i=0; i<M; i++)//建图
        {
            scanf("%s%lf%s",str_1,&num,str_2);
            int from,to;
            for(int j=1; j<=N; j++)
            {
                if(strcmp(str[j],str_1)==0)
                    from=j;
                if(strcmp(str[j],str_2)==0)//坑点就在这里
                    to=j;//to和from可能是同一点
            }
            map[from][to]=num;//建图
        }
        for(int k=1; k<=N; k++)
            for(int i=1; i<=N; i++)
                for(int j=1; j<=N; j++)
                    if(map[i][j]<map[i][k]*map[k][j])
                        map[i][j]=map[i][k]*map[k][j];
        bool flag=0;
        for(int i=1; i<=N; i++)
            if(map[i][i]>1.0)
            {
                flag=1;
                break;
            }
        flag==1?printf("Case %d: Yes\n",casen++):printf("Case %d: No\n",casen++);
//        getchar();
    }
    return 0;
}

讲道理还可以用贝尔曼-福特，SPFA各做一遍

贝尔曼-福特

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<time.h>
using namespace std;
//贝尔曼-福特
const int maxn=1111;
int u[maxn];
int v[maxn];
double w[maxn];
double dis[maxn];
char str[33][33];
int N;//点的数量
int M;//边的数量
void init_map()//建图
{
    for(int i=0; i<N; i++) //N个顶点
        scanf("%s",str[i]);
    scanf("%d",&M);
    char str_1[33],str_2[33];
    double num;
    for(int i=0; i<M; i++)
    {
        scanf("%s%lf%s",str_1,&num,str_2);
        int from,to;
        for(int j=0; j<N; j++)
        {
            if(strcmp(str_1,str[j])==0)
                from=j;
            if(strcmp(str_2,str[j])==0)
                to=j;
        }
        u[i]=from;
        v[i]=to;
        w[i]=num;
    }
}
bool Bellman_Ford(int star)
{
    memset(dis,0,sizeof(dis));
    dis[star]=1.0;
    for(int k=1; k<=N; k++) //由于要形成环，所以要循环N次
    {
        for(int i=0; i<M; i++)
        {
            if(dis[v[i]]<dis[u[i]]*w[i])
            {
                dis[v[i]]=dis[u[i]]*w[i];
            }
        }
    }
    if(dis[star]>1.0)
        return true;
    return false;
}
int main()
{
    int casen=1;
    while(~scanf("%d",&N)&&N)
    {
        init_map();
        bool flag=0;
        for(int i=0; i<N; i++)
            if(Bellman_Ford(i)==true)
            {
                flag=1;
                break;
            }
        //等待输出
        flag==1?printf("Case %d: Yes\n",casen++):printf("Case %d: No\n",casen++);
    }
    return 0;
}