链接:https://vjudge.net/problem/POJ-1860
题意:
有N个点,支持货币兑换,从货币a->b手续费c,汇率r。
求能否换一圈使总净额增加。
思路:
bellman-ford。
找一个正权回路。
代码:
#include <iostream>
#include <memory.h>
using namespace std;
const int MAXN = 210;
double dis[MAXN];
int n,m,s;
double v;
int w = 1;
struct Node
{
int a,b;
double r,c;
}node[210];
bool bellman_Ford()
{
memset(dis,0, sizeof(dis));
dis[s] = v;
for (int i = 1;i <= n-1;i++)
{
bool flag = false;
for (int j = 1;j<w;j++)
{
int a = node[j].a,b = node[j].b;
double r = node[j].r,c = node[j].c;
if (dis[b] < (dis[a]-c)*r)
{
dis[b] = (dis[a]-c)*r;
flag = true;
}
}
if (!flag)
break;
}
for (int i = 1;i < w;i++)
if (dis[node[i].b] < (dis[node[i].a] - node[i].c)*node[i].r)
return true;
return false;
}
int main()
{
int a,b;
double rab,cab,rba,cba;
scanf("%d%d%d%lf",&n,&m,&s,&v);
for (int i = 0;i<m;i++)
{
scanf("%d%d%lf%lf%lf%lf",&a,&b,&rab,&cab,&rba,&cba);
node[w].a = a;
node[w].b = b;
node[w].r = rab;
node[w++].c = cab;
node[w].b = a;
node[w].a = b;
node[w].r = rba;
node[w++].c = cba;
}
if (bellman_Ford())
printf("YES
");
else
printf("NO
");
return 0;
}