poj 1860 Currency Exchange :bellman-ford
2013-08-20 18:18
295 查看
点击打开链接
Currency Exchange
Description
Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the same pair of currencies.
Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency.
For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR.
You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges,
and real RAB, CAB, RBA and CBA - exchange rates and commissions when exchanging A to B and B to A respectively.
Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative
sum of money while making his operations.
Input
The first line of the input contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description
of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1<=S<=N<=100, 1<=M<=100, V is real number, 0<=V<=103.
For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10-2<=rate<=102, 0<=commission<=102.
Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations
will be less than 104.
Output
If Nick can increase his wealth, output YES, in other case output NO to the output file.
Sample Input
Sample Output
bellman-ford算法,题目大意是有N种货币和M个货币交换点,Nick有一些其中一种货币,每两种货币兑换有一个公式,就是(本金 - 手续费) * 转换率,每个交换点只能交换某两种特定的货币,最后问是否可以通过这些交换点使得最后的本金会增加
bellman-ford计算是否有负圈回路就好,其实就是判断是否有使本金增长的圈
#include<stdio.h>
double map[101][101][2];
double dis[101];
int n, m, s;
double v;
bool bellman()
{
int i, j, k;
for(i = 1; i < 101; i++)
{
dis[i] = 0;
}
dis[s] = v;
for(i = 1; i < n; i++)
{
for(j = 1; j <= n; j++)
{
for(k = 1; k <= n; k++)
{
if(map[j][k][0] > 0)
{
if(dis[k] < (dis[j] - map[j][k][1]) * map[j][k][0])
{
dis[k] = (dis[j] - map[j][k][1]) * map[j][k][0];
}
}
}
}
}
for(j = 1; j <= n; j ++)
{
for(k = 1; k <= n; k++)
{
if(dis[k] < (dis[j] - map[j][k][1]) * map[j][k][0])
return 0;
}
}
return 1;
}
int main()
{
scanf("%d %d %d %lf", &n, &m, &s, &v);
int i = m;
int a, b;
while(i--)
{
scanf("%d %d", &a, &b);
scanf("%lf %lf %lf %lf", &map[a][b][0], &map[a][b][1], &map[b][a][0], &map[b][a][1]);
}
if(bellman() == 0)
printf("YES\n");
else
printf("NO\n");
return 0;
}
Currency Exchange
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 16635 | Accepted: 5821 |
Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the same pair of currencies.
Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency.
For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR.
You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges,
and real RAB, CAB, RBA and CBA - exchange rates and commissions when exchanging A to B and B to A respectively.
Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative
sum of money while making his operations.
Input
The first line of the input contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description
of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1<=S<=N<=100, 1<=M<=100, V is real number, 0<=V<=103.
For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10-2<=rate<=102, 0<=commission<=102.
Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations
will be less than 104.
Output
If Nick can increase his wealth, output YES, in other case output NO to the output file.
Sample Input
3 2 1 20.0 1 2 1.00 1.00 1.00 1.00 2 3 1.10 1.00 1.10 1.00
Sample Output
YES
bellman-ford算法,题目大意是有N种货币和M个货币交换点,Nick有一些其中一种货币,每两种货币兑换有一个公式,就是(本金 - 手续费) * 转换率,每个交换点只能交换某两种特定的货币,最后问是否可以通过这些交换点使得最后的本金会增加
bellman-ford计算是否有负圈回路就好,其实就是判断是否有使本金增长的圈
#include<stdio.h>
double map[101][101][2];
double dis[101];
int n, m, s;
double v;
bool bellman()
{
int i, j, k;
for(i = 1; i < 101; i++)
{
dis[i] = 0;
}
dis[s] = v;
for(i = 1; i < n; i++)
{
for(j = 1; j <= n; j++)
{
for(k = 1; k <= n; k++)
{
if(map[j][k][0] > 0)
{
if(dis[k] < (dis[j] - map[j][k][1]) * map[j][k][0])
{
dis[k] = (dis[j] - map[j][k][1]) * map[j][k][0];
}
}
}
}
}
for(j = 1; j <= n; j ++)
{
for(k = 1; k <= n; k++)
{
if(dis[k] < (dis[j] - map[j][k][1]) * map[j][k][0])
return 0;
}
}
return 1;
}
int main()
{
scanf("%d %d %d %lf", &n, &m, &s, &v);
int i = m;
int a, b;
while(i--)
{
scanf("%d %d", &a, &b);
scanf("%lf %lf %lf %lf", &map[a][b][0], &map[a][b][1], &map[b][a][0], &map[b][a][1]);
}
if(bellman() == 0)
printf("YES\n");
else
printf("NO\n");
return 0;
}
相关文章推荐
- POJ 1860 Currency Exchange (Bellman-Ford 找正环)
- POJ 1860 反向Bellman-Ford
- 【Bellman_Ford】poj 1860 Currency Exchange
- 【bellman-Ford判断正权回路】POJ - 1860 Currency Exchange
- poj 1860 (Bellman_Ford判断正环)
- POJ1860-Currency Exchange(bellman-ford求正环)
- Bellman_Ford变形求最长路+正权回路或spfa——POJ 1860
- POJ 1860 bellman_ford
- POJ 1860 bellman-ford 的变形
- poj 1860 Bellman-ford 逆向思维
- Currency Exchange POJ - 1860 Bellman-Ford最短路
- POJ 1860 Currency Exchange (Bellman-Ford)
- POJ 1860 Currency Exchange 毫无优化的bellman_ford跑了16Ms,spfa老是WA。。
- POJ 1860 Currency Exchange(Bellman-Ford)
- POJ 1860 Currency Exchange (Bellman-Ford)
- poj 1860 Currency Exchange (bellman-ford 判断正环)
- poj 1860 Bellman-Ford
- poj 1860 Currency Exchange (最短路bellman_ford思想找正权环 最长路)
- POJ1860 换零钱套利 图论(Bellman-ford)
- POJ 1860 Currency Exchange(Bellman-Ford)