poj 1860

题目

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<=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 20.01 2 1.00 1.00 1.00 1.002 3 1.10 1.00 1.10 1.00

Sample Output

YES

题目要求就是每一行给出两个点，后面跟着四个浮点数代表a到b的转化率，a到b的手续费，b到a的转化率，b到a的手续费

一开始把手续费当作百分比的，然后wa了= =，后来才行知道不管转化多少钱都得交这些手续费。。。

这个题是用spfa算法做的，利用队列进行更新dist

这是自己手写的第一个spfa算法题，不是转载，纪念纪念- -，

代码如下，为了避免看我代码看不懂我做了注释

//728K63MS#include<iostream>#include<vector>#include<cstdio>#include<queue>using namespace std;struct lu{    int endd;//连接的点    float r,c;//r是转化率，c是手续费    lu(int a,float rr,float cc){endd=a;r=rr;c=cc;}};float dist[105]={0};//将距离理解为钱就好，初始只有某种钱（x）多少元，其他位置都为0，最后比较x与初始钱数的大小来判断是否有正环vector<lu>tu[105];int n,m,v;float q;queue<int>dui;bool vis[105]={0};//判断当前队列中是否有某点bool spfa(int v){    float y=dist[v];    dui.push(v);    vis[v]=!vis[v];    while(!dui.empty())    {    int now=dui.front();dui.pop();    for(int i=0;i<tu[now].size();++i)    {        int ed=tu[now][i].endd;        if(dist[ed]<(dist[now]-tu[now][i].c)*tu[now][i].r)//如果这个点的钱数可以变成更大值，就进行更新        {            dist[ed]=(dist[now]-tu[now][i].c)*tu[now][i].r;            if(!vis[ed]){dui.push(ed);vis[ed]=1;}//如果当前队列没有这个数，就把这个数加入队列，用这个数来对其他种类的钱数进行松弛        }    }    vis[now]=!vis[now];//这个点操作完了，也pop了，就把vis进行标记为不在队列    }    if(dist[v]>y)return true;//最后判断一开始手里面的那种钱比最后多还是少，多就是存在正环    else return false;}int main(){    int a,b;float rab,cab,rba,cba;    scanf("%d%d%d%f",&n,&m,&v,&q);    dist[v]=q;    for(int i=0;i<m;++i)    {        scanf("%d%d%f%f%f%f",&a,&b,&rab,&cab,&rba,&cba);        tu[a].push_back(lu(b,rab,cab));        tu[b].push_back(lu(a,rba,cba));    }    bool flag=spfa(v);    if(flag)cout<<"YES"<<endl;    else cout<<"NO"<<endl;}

下面这个方法是上面这个程序的微改，速度由64MS变成了16MS
调整的地方在于找到自增环的时候自动退出程序...
这是我刚刚出去散步考虑的一种情况，如果a到b的手续费为0.01，转化率为2，同理b到a也一样，这样就会陷入一个循环，一直循环而找不到最大值
所以在第一次找到自增环就可以退出程序了。
代码如下

#include<iostream>#include<vector>#include<cstdio>#include<queue>using namespace std;struct lu{    int endd;    float r,c;    lu(int a,float rr,float cc){endd=a;r=rr;c=cc;}};float dist[105]={0};vector<lu>tu[105];int n,m,v;float q;queue<int>dui;bool vis[105]={0};bool bellman_ford(int v){    dui.push(v);    vis[v]=!vis[v];    while(!dui.empty())    {    int now=dui.front();dui.pop();    for(int i=0;i<tu[now].size();++i)    {        int ed=tu[now][i].endd;        if(dist[ed]<(dist[now]-tu[now][i].c)*tu[now][i].r)        {            dist[ed]=(dist[now]-tu[now][i].c)*tu[now][i].r;            if(ed==v)return true;            if(!vis[ed]){dui.push(ed);vis[ed]=1;}        }    }    vis[now]=!vis[now];    }    return false;}int main(){    int a,b;float rab,cab,rba,cba;    scanf("%d%d%d%f",&n,&m,&v,&q);    dist[v]=q;    for(int i=0;i<m;++i)    {        scanf("%d%d%f%f%f%f",&a,&b,&rab,&cab,&rba,&cba);        tu[a].push_back(lu(b,rab,cab));        tu[b].push_back(lu(a,rba,cba));    }    bool flag=bellman_ford(v);    if(flag)cout<<"YES"<<endl;    else cout<<"NO"<<endl;}