POJ 1860：Currency Exchange：bellman最短路变形进行盈利判环

Currency Exchange

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 19064		Accepted: 6814

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 R _AB, C _AB, R _BA and C _BA - 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<=10 ³. 
For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10 ^-2<=rate<=10 ², 0<=commission<=10 ². 
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 10 ⁴. 

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

Source

Northeastern Europe 2001, Northern Subregion

图论的入门，判断存不存在盈利的环。递归搜索。简答的暴力搜索。

该题目的标准解法是bellman_ford的最短路变形，较有难度。原来的算法是，如果有负权回路，则最短路的n-1次松弛后不收敛。这道题的情况刚好相反，回路的值逐圈增加，那么使用bellman_ford算法求解最长路径，最长路的价值在n-1次松弛后也不收敛，从而判断出有盈利回路。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int n,m,s,ta,tb;
double v,r[105][105],c[105][105],use[105];
int result;
void init()
{
	int i,j;
	for(i=0;i<105;i++)
	{
		for(j=0;j<105;j++)
			r[i][j]=c[i][j]=-1;
		use[i]=-1;
	}
	for(i=1;i<=m;i++)
	{
		scanf("%d%d%",&ta,&tb);
		scanf("%lf%lf%lf%lf",&r[ta][tb],&c[ta][tb],&r[tb][ta],&c[tb][ta]);
	}
	result=0;
}
void search(int x,double y)
{
	int i,j;
	if(result||y<=0)
		return ;
	if(use[x]!=-1)
	{
		if(y>use[x])
		{
			result=1;
			return ;
		}
	}
	else
	{
		use[x]=y;
		for(i=1;i<=n;i++)
		{
			if(r[x][i]!=-1)
				search(i,(use[x]-c[x][i])*r[x][i]);
		}
		use[x]=-1;
	}
}
int main()
{
	int i,j;
	while(scanf("%d%d%d%lf",&n,&m,&s,&v)!=EOF)
	{
		init();
		search(s,v);
		if(result)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}

bellman_ford改进算法的实现：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int n,m,s,ta,tb;
double v,r[105][105],c[105][105],use[105],d[105];
int result;
void init()
{
	int i,j;
	for(i=0;i<105;i++)
	{
		for(j=0;j<105;j++)
			r[i][j]=c[i][j]=-1;
		use[i]=-1;
		d[i]=0;
	}
	for(i=1;i<=m;i++)
	{
		scanf("%d%d%",&ta,&tb);
		scanf("%lf%lf%lf%lf",&r[ta][tb],&c[ta][tb],&r[tb][ta],&c[tb][ta]);
	}
	result=0;
}
void bellman()
{
	int i,j,k;
	d[s]=v;
	for(k=0;k<n-1;k++)
	{
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				if(r[i][j]!=-1&&d[i]>c[i][j]&&d[j]<(d[i]-c[i][j])*r[i][j])
					d[j]=(d[i]-c[i][j])*r[i][j];
			}
		}
	}
	for(i=1;i<=n;i++)
	{
		if(result)
			break;
		for(j=1;j<=n;j++)
		{
			if(r[i][j]!=-1&&d[i]>c[i][j]&&d[j]<(d[i]-c[i][j])*r[i][j])
			{
				result=1;
				break;
			}
		}
	}
}
int main()
{
	int i,j;
	while(scanf("%d%d%d%lf",&n,&m,&s,&v)!=EOF)
	{
		init();
		bellman();
		if(result)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}