Bellman 算法

这道题目事实上就是在求有没有正环。与求负环的差别就是要不断的更新值，可是这个值要变大。而不是变小。

Currency Exchange

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 20441		Accepted: 7337

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 20.0
1 2 1.00 1.00 1.00 1.00
2 3 1.10 1.00 1.10 1.00

Sample Output

YES

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <stack>
#include <deque>
#include <vector>
#include <set>
#include <map>
using namespace std;
const int INF = 99999999.9;
const int EM = 5555;
const int VM = 110;
int n,m,s;
double mon;

struct Edge{
    int u,v;
    double com;
    double exg;
}edge[EM<<1];
int cnt;
double dis[VM];
//double mon;

/*void addage(int cu,int vu,double aa,double bb){
    edge[cnt].from = cu;
    edge[cnt].to = vu;
    edge[cnt].com = aa;
    edge[cnt].exg = bb;
    cnt++;
}*/

int Bellman(){
    int i,j,flag;
    for(i=1;i<=n;i++){
        dis[i] = 0.0;
    }
    dis[s] = mon;
    for(i=1;i<n;i++){
        bool flag = false;
        for(j=0;j<cnt;j++){
            int u = edge[j].u;
            int v = edge[j].v;
            double rate = edge[j].exg;
            double cost = edge[j].com;
            if(dis[v]<(dis[u]-cost)*rate)//求最大的路径
            {
                dis[v]=(dis[u]-cost)*rate;
                flag=true;
            }
            //if(dis[edge[j].to] < (dis[edge[j].from]-edge[j].com)*edge[j].exg){
            //    dis[edge[j].to] = (dis[edge[j].from]-edge[j].com)*edge[j].exg;
            //    flag = 1;
            //}
        }
        if(flag==false){
            return false;
        }
    }
    for(j=0; j<cnt; j++)
    {
        if(dis[edge[j].v]<(dis[edge[j].u]-edge[j].com)*edge[j].exg)//与传统的bell不一样，传统的bell是找负环。如今是找正环，正环无限松弛
            return true;
    }
    return false;
}

int main(){
    //int n,m,s;
    //double mon;
    double rab,cab,rba,cba;
    int marka,markb;

    while(~scanf("%d%d%d%lf",&n,&m,&s,&mon)){
        cnt = 0;
        //flag = 1;
        while(m--){
            scanf("%d%d%lf%lf%lf%lf",&marka,&markb,&rab,&cab,&rba,&cba);
            //addage(marka,markb,a2,a1);
            //addage(markb,marka,b2,b1);
            edge[cnt].u=marka,edge[cnt].v=markb,edge[cnt].com=cab,edge[cnt].exg=rab;
            cnt++;
            edge[cnt].u=markb,edge[cnt].v=marka,edge[cnt].com=cba,edge[cnt].exg=rba;
            cnt++;
        }
        if(Bellman()){
            printf("YES\n");
        }
        else{
            printf("NO\n");
        }
    }

    return 0;
}