C - C(换钱问题)

换钱问题：

给出n种钱，m个站点，现在有第 s种钱，身上有v 这么多；

下面 m行 站点有a,b两种钱，rab a->b的汇率，cab a-->b的手续费， 相反rba cba ； 

问在10次之内，能不能把钱换的变的多了？

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<=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

方法1：用Bellman算法，若存在正权回路，则说明货币经转化是越来越大的，即满足题意。

　　　  所以，这道题就是在让你求回路中是否存在正权回路。

#include "stdio.h"
#include "string.h"
#include "algorithm"
using namespace std;
int n,m,x,flag,u[220],v[220];
double y,dis[220],r1[220],c1[220],r2[220],c2[220];
void djk()
{
    for(int i=1;i<n;i++)
    {
        int ans=0;
        for(int j=1;j<=m;j++)
        {
            if(dis[v[j]]<(dis[u[j]]-c1[j])*r1[j])
            {
                dis[v[j]]=(dis[u[j]]-c1[j])*r1[j];
                ans++;
            }
            if(dis[u[j]]<(dis[v[j]]-c2[j])*r2[j])
            {
                dis[u[j]]=(dis[v[j]]-c2[j])*r2[j];
                ans++;
            }
         }
         if(ans==0) break;
    }
    flag=0;
    for(int j=1;j<=m;j++)
    {
        if(dis[v[j]]<(dis[u[j]]-c1[j])*r1[j])
        {
            flag=1;
            break;
        }
        if(dis[u[j]]<(dis[v[j]]-c2[j])*r2[j])
        {
            flag=1;
            break;
        }
    }
}
int main()
{
    while(~scanf("%d%d%d%lf",&n,&m,&x,&y))
    {
        memset(dis,0,sizeof(dis));
        dis[x]=y;
        for(int i=1;i<=m;i++)
        scanf("%d%d%lf%lf%lf%lf",&u[i],&v[i],&r1[i],&c1[i],&r2[i],&c2[i]);
        djk();
        if(flag) printf("YES\n");
        else printf("NO\n");
    }
}

方法2：

　　SPFA方法：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#define ll long long
#define INF 0x3f3f3f
using namespace std;
typedef pair<int,int> P;
const int maxn = 1e2+10;
int n,m,s,tot;
double v;
double dis[maxn];
int head[maxn],vis[maxn];
double cost[maxn][maxn],rate[maxn][maxn];
struct node
{
    int to;
    double w;
    int next;
} edge[maxn*maxn];

void add(int u,int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}

bool SPFA(int s,double v)
{
    for(int i=1; i<=n; i++)
    {
        dis[i] = 0;
        vis[i] = 0;
    }
    dis[s] = v;
    vis[s] = 1;
    queue<int> q;
    q.push(s);
    while(!q.empty())
    {
        int x = q.front();
        q.pop();
        vis[x] = 0;
        for(int i=head[x]; i!=-1; i=edge[i].next)
        {
            int j = edge[i].to;
            if(dis[j]<(dis[x]-cost[x][j])*rate[x][j])
            {
                dis[j] = (dis[x]-cost[x][j])*rate[x][j];
                if(dis[s]>v)
                    return true;
                if(!vis[j])
                {
                    q.push(j);
                    vis[j] = 1;
                }
            }
        }
    }
    return false;
}

int main()
{
    while(cin>>n>>m>>s>>v)
    {
        int i;
        memset(cost,0,sizeof(cost));
        memset(rate,0,sizeof(rate));
        memset(head,-1,sizeof(head));
        tot = 1;
        for(i=0; i<m; i++)
        {
            int a,b;
            double cab,rab,cba,rba;
            scanf("%d%d%lf%lf%lf%lf",&a,&b,&rab,&cab,&rba,&cba);
            cost[a][b] = cab;
            cost[b][a] = cba;
            rate[a][b] = rab;
            rate[b][a] = rba;
            add(a,b);
            add(b,a);
        }
        if(SPFA(s,v))
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}