poj - 1860 Currency Exchange Bellman-Ford 判断正环

Currency Exchange POJ - 1860

题意：

　　有许多货币兑换点，每个兑换点仅支持两种货币的兑换，兑换有相应的汇率和手续费。你有s这个货币 V 个，问是否能通过合理地兑换货币，使得你手中的货币折合成s后是有增加的。

思路：

　　这道题在建立每种货币的兑换关系后，找到图中的正环即可，因为你沿着正环跑就可以增加价值。这里可以用类似Bellman_Ford判断负环的方法。

#include <algorithm>

#include  <iterator>

#include  <iostream>

#include   <cstring>

#include   <cstdlib>

#include   <iomanip>

#include    <bitset>

#include    <cctype>

#include    <cstdio>

#include    <string>

#include    <vector>

#include     <stack>

#include     <cmath>

#include     <queue>

#include      <list>

#include       <map>

#include       <set>

#include   <cassert>

using namespace std;

//#pragma GCC optimize(3)

//#pragma comment(linker, "/STACK:102400000,102400000")  //c++

#define lson (l , mid , rt << 1)

#define rson (mid + 1 , r , rt << 1 | 1)

#define debug(x) cerr << #x << " = " << x << "\n";

#define pb push_back

#define pq priority_queue

typedef long long ll;

typedef unsigned long long ull;

typedef pair<ll ,ll > pll;

typedef pair<int ,int > pii;

typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q

//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q

#define fi first

#define se second

//#define endl '\n'

#define OKC ios::sync_with_stdio(false);cin.tie(0)

#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行

#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)

//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //

const ll nmos = 0x80000000;  //-2147483648

const int inf = 0x3f3f3f3f;

const ll inff = 0x3f3f3f3f3f3f3f3f; //

const int mod = 1e9+;

const double esp = 1e-;

const double PI=acos(-1.0);

template<typename T>

inline T read(T&x){

    x=;int f=;char ch=getchar();

    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();

    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();

    return x=f?-x:x;

}

/*-----------------------showtime----------------------*/

                const int maxn = ;

                struct edge

                {

                        int u,v;

                        double c,r;

                }e[maxn];

                int n,m,s,tot = ;

                bool flag;

                double val,dis[maxn];

                void add(int u,int v,double c,double r){

                    e[++tot].u = u;

                    e[tot].v = v;

                    e[tot].c = c;

                    e[tot].r = r;

                }

                double get(int i){

                    return 1.0*(dis[e[i].u] - e[i].c) * e[i].r;

                }

                bool bellman_ford(){

                    for(int i=; i<=n; i++)dis[i] = 0.0;

                    dis[s] = val;

                    for(int i=; i<n; i++){

                        flag = true;

                        for(int j=; j<=tot; j++){

                            if(dis[e[j].v] < get(j))

                            {

                                dis[e[j].v] = get(j);

                                flag = false;

                            }

                        }

                        if(dis[s] > val)return true;

                        if(flag) break;

                    }

                    for(int i=; i<=tot; i++){

                        if(dis[e[i].v] < get(i)){

                            return true;

                        }

                    }

                    return false;

                }

int main(){

                scanf("%d%d%d%lf", &n, &m, &s, &val);

                int u,v;double c1,r1,c2,r2;

                for(int i=; i<=m; i++){

                    scanf("%d%d%lf%lf%lf%lf",&u,&v,&r1,&c1,&r2,&c2);

                    add(u,v,c1,r1);

                    add(v,u,c2,r2);

                }

                if(bellman_ford())puts("YES");

                else puts("NO");

                return ;

}

POJ - 1860