POJ 1860 Currency Exchange

Test instructions: There are n kinds of currency, can be exchanged with each other, there are M redemption rules, exchange rules give the exchange rate R and handling fee C, the formula is B = (a-c) * R, from a currency to B currency, ask whether you can make money through constant conversion, the amount in the hands of the exchange cannot be negative.

Solution: Bellman-ford. Map: The currency as a point, each redemption rule for the edge, two points of the path length of the money after the exchange. After the map can be seen test instructions for the existence of positive ring in the figure, with Bellman-ford to find the longest path, if there is positive ring output Yes, there is no output.

Code:

#include <stdio.h> #include <iostream> #include <algorithm> #include <string> #include < string.h> #include <math.h> #include <limits.h> #include <time.h> #include <stdlib.h># include<map> #include <queue> #include <set> #include <stack> #include <vector> #include <iomanip> #define LL Long Long#define Lson L, M, RT << 1#define Rson m + 1, R, RT << 1 |    1using namespace Std;struct node{int u, v;    Double R, C; node (int u, int v, double R, double C): U (U), V (v), R (R), C (c) {} node () {}}edge[205];int n, M, St, cnt;double money;d        Ouble Dis[105];bool Bellmanford () {for (int i = 1; I <= n; i++) {if (i = = st) Dis[i] = money; else Dis[i] = 0;//is initialized to 0 because there can be no negative amount in the process} for (int i = 1; i < n; i++)//n-1 times relaxation for (int j = 0; J < CNT;    ) DIS[EDGE[J].V] = max ((dis[edge[j].u]-edge[j].c) * EDGE[J].R, DIS[EDGE[J].V]); for (int i = 0; i < cnt; i++)//If you can relax after n-1, sayThe existence of positive ring if (DIS[EDGE[I].V] < (DIS[EDGE[I].U]-edge[i].c) * EDGE[I].R) return 1; return 0;}        int main () {while (~scanf ("%d%d%d%lf", &n, &m, &st, &money)) {cnt = 0;            for (int i = 0; i < m; i++) {int u, v;            Double R1, C1, R2, C2;            scanf ("%d%d%lf%lf%lf%lf", &u, &v, &r1, &c1, &AMP;R2, &AMP;C2);            edge[cnt++] = node (u, V, R1, C1);        edge[cnt++] = node (V, u, R2, C2);        } if (Bellmanford ()) puts ("YES");    Else puts ("NO"); } return 0;}

　　

POJ 1860 Currency Exchange