最短路径练习-POJ 1860 Currency Exchange

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题目大意：

多种汇币，汇币之间可以无限转换（存在手续费），问是否存在一种路径使金钱一直增加。

思路：

判断是否存在正权回路，用belman-Ford算法的变体更改松弛条件。

#include<stdio.h>#include<stdlib.h>#include <string.h>#define MAXINT 1100#define MAXD 999999struct edge{int u, v;double r, c;}ed[MAXINT];bool bellman_ford(int n, int v, double c, int e){double d[MAXINT];memset(d, 0, sizeof(d));d[v] = c;bool flag;for (int i = 1; i < n; i++){flag = false;for (int j = 0; j < e; j++)if (d[ed[j].v] < (d[ed[j].u] - ed[j].c)*ed[j].r){d[ed[j].v] = (d[ed[j].u] - ed[j].c)*ed[j].r;flag = true;}if (!flag)break;}for (int i = 0; i < e; i++)if (d[ed[i].v] < (d[ed[i].u] - ed[i].c)*ed[i].r)return true;return false;}int main(){int n, m, s;int x, y, e;double v, r1, c1, r2, c2;while (scanf("%d%d%d%lf", &n, &m, &s, &v) != EOF){e = 0;for (int i = 1; i <= m; i++){scanf("%d%d%lf%lf%lf%lf", &x, &y, &r1, &c1, &r2, &c2);ed[e].u = x;ed[e].v = y;ed[e].r = r1;ed[e++].c = c1;ed[e].u = y;ed[e].v = x;ed[e].r = r2;ed[e++].c = c2;}if (bellman_ford(n, s, v, e))printf("YES\n");elseprintf("NO\n");}return 0;}

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