POJ - 1860 Currency Exchange（反用bellman求正环）

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

题意：这个题太长辣，简单来说就是求一个图中是否存在正环。

思路：可以想到bellman算法是可以找出是否存在负环的，那么我们就可以通过反向使用这儿算法来求是否存在正环。

但是要注意的是，不能仅仅通过更新v-1次来判断，因为题目要求的是要包含起点的正圈。

代码：

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <numeric>
#include <set>
#include <string>
#include <cctype>
#include <sstream>
#define INF 0x3f3f3f3f
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
using namespace std;
typedef long long LL;
typedef pair<LL, LL> P;
const int maxn = 1e2 + 5;
const int mod = 1e8 + 7;

int n,m,s;
double v;
double d[maxn];
struct node{
    int a,b;
    double r,c;
}G[maxn];
bool bellman(int t){
    memset(d,0,sizeof(d));
    d[s]=v;
    for (int i=0;i<n-1;i++){
        bool flag=0;
        for (int j=0;j<t;j++){
            if (d[G[j].b]<(d[G[j].a]-G[j].c)*G[j].r){
                d[G[j].b]=(d[G[j].a]-G[j].c)*G[j].r;
                flag=1;
            }
        }
        if (!flag) break;   //结束松弛
    }
    for (int i=0;i<t;i++){  //判断从s开始的点时候能无限松弛，因为存在正环的话才能无限松弛
        if (d[G[i].b]<(d[G[i].a]-G[i].c)*G[i].r) return 1;
    }
    return 0;
}
int main() {
    //freopen ("in.txt", "r", stdin);
    int a,b;
    double rab,cab,rba,cba;
    while(~scanf ("%d%d%d%lf",&n,&m,&s,&v)){
        int t=0; //总边数
        for (int i=0;i<m;i++){
            scanf ("%d%d%lf%lf%lf%lf",&a,&b,&rab,&cab,&rba,&cba);
            G[t].a=a;
            G[t].b=b;
            G[t].r=rab;
            G[t++].c=cab;
            G[t].a=b;
            G[t].b=a;
            G[t].r=rba;
            G[t++].c=cba;
        }
        if (bellman(t)) printf ("YES\n");
        else printf ("NO\n");
    }
    return 0;
}