zoj1092

maksyuki 发表于 oj 分类，标签: 图论-最短路

11 9月 2015

Arbitrage

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 Specification

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 c_i of a source currency, a real number r_ij which represents the exchange rate from c_ito c_j and a name c_j 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 Specification

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

3USDollarBritishPoundFrenchFranc3USDollar 0.5 BritishPoundBritishPound 10.0 FrenchFrancFrenchFranc 0.21 USDollar 3USDollarBritishPoundFrenchFranc6USDollar 0.5 BritishPoundUSDollar 4.9 FrenchFrancBritishPound 10.0 FrenchFrancBritishPound 1.99 USDollarFrenchFranc 0.09 BritishPoundFrenchFranc 0.19 USDollar 0

Sample Output

Case 1: YesCase 2: No

题目类型：最长回路

算法分析：调用n次bellman-ford算法计算每一点为源点的最长回路(乘积)，对于长度大于1.0的情况直接可以判断出出现套汇

#include <iostream>
#include <fstream>
#include <algorithm>
#include <iomanip>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <map>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <list>
#include <ctime>
using namespace std;

const int maxn = 36 + 6;
const double EPS = 1e-6;


struct Node
{
    int u, v;
    double w;
};

Node edge[maxn*maxn];

char input[maxn][maxn];
double dis[maxn];

bool is_valid;
int n, flag = 1, m;

void bellman_ford (int s)
{
    memset (dis, 0, sizeof (dis));
    dis[s] = 1;

    int i;
    for (i = 1; i <= n; i++)
    {
        int j;
        for (j = 0; j < m; j++)
        {
            if (dis[edge[j].u] * edge[j].w > dis[edge[j].v] + EPS)
                dis[edge[j].v] = dis[edge[j].u] * edge[j].w;
        }
    }

    if (dis[s] > 1.0 + EPS)
        is_valid = true;

}

int main()
{
//    freopen ("aaa.txt", "r", stdin);

    while (scanf ("%d", &n) != EOF)
    {
        if (n == 0)
            break;

        int i;
        for (i = 0; i < n; i++)
            scanf ("%s", input[i]);

        scanf ("%d", &m);

        for (i = 0; i < m; i++)
        {
            char temp_a[maxn], temp_b[maxn];
            double temp_w;
            scanf ("%s %lf %s", temp_a, &temp_w, temp_b);

            int j;
            for (j = 0; strcmp (temp_a, input[j]); j++) ;
            edge[i].u = j;

            for (j = 0; strcmp (temp_b, input[j]); j++) ;
            edge[i].v = j;

            edge[i].w = temp_w;
        }

        is_valid = false;

        for (i = 0; i < n; i++)
        {
            bellman_ford (i);

            if (is_valid)
                break;
        }

        cout << "Case " << flag++ << ": ";

        if (is_valid)
            cout << "Yes" << endl;

        else
            cout << "No" << endl;
    }

return 0;
}

100

101

102

103

104

105

106

107

108

#include <iostream>

#include <fstream>

#include <algorithm>

#include <iomanip>

#include <cstring>

#include <cstdio>

#include <cmath>

#include <map>

#include <string>

#include <vector>

#include <stack>

#include <queue>

#include <set>

#include <list>

#include <ctime>

using namespace std;

const int maxn = 36 + 6;

const double EPS = 1e-6;

struct Node

{

int u, v;

double w;

};

Node edge[maxn*maxn];

char input[maxn][maxn];

double dis[maxn];

bool is_valid;

int n, flag = 1, m;

void bellman_ford (int s)

{

memset (dis, 0, sizeof (dis));

dis[s] = 1;

int i;

for (i = 1; i <= n; i++)

{

int j;

for (j = 0; j < m; j++)

{

if (dis[edge[j].u] * edge[j].w > dis[edge[j].v] + EPS)

dis[edge[j].v] = dis[edge[j].u] * edge[j].w;

}

if (dis[s] > 1.0 + EPS)

is_valid = true;

}

int main()

{

// freopen ("aaa.txt", "r", stdin);

while (scanf ("%d", &n) != EOF)

{

if (n == 0)

break;

int i;

for (i = 0; i < n; i++)

scanf ("%s", input[i]);

scanf ("%d", &m);

for (i = 0; i < m; i++)

{

char temp_a[maxn], temp_b[maxn];

double temp_w;

scanf ("%s %lf %s", temp_a, &temp_w, temp_b);

int j;

for (j = 0; strcmp (temp_a, input[j]); j++) ;

edge[i].u = j;

for (j = 0; strcmp (temp_b, input[j]); j++) ;

edge[i].v = j;

edge[i].w = temp_w;

}

is_valid = false;

for (i = 0; i < n; i++)

{

bellman_ford (i);

if (is_valid)

break;

}

cout << "Case " << flag++ << ": ";

if (is_valid)

cout << "Yes" << endl;

else

cout << "No" << endl;

}

return 0;

}

zoj1091

zoj1203

一	二	三	四	五	六	日
« 6月
						1
2	3	4	5	6	7	8
9	10	11	12	13	14	15
16	17	18	19	20	21	22
23	24	25	26	27	28	29
30	31

超客的水墨烟雨

zoj1092

Input Specification

Output Specification

Sample Input

Sample Output

分类目录

标签云

书签

近期评论