POJ1860-Currency Exchange

这道题讲述了城里有几个点能过换钱问能否套利。

只要用Bellmen_Ford判断图中是否含有负圈即可。

#include <cstdio>
#include <cstring>

const int maxn = 100 + 5;
const int maxm = 100 + 5;

struct Edge {
int from;
int to;
double rate;
double com;
};

Edge es[2 * maxm];
double d[maxn];

bool find_negative_loop(int s, int v, int l, double sum) {
memset(d, sizeof(d), 0);
d[s] = sum;
for (int i = 0; i < v; i++) {
for (int j = 0; j < l; j++) {
Edge e = es[j];
if (d[e.to] < (d[e.from] - e.com) * e.rate) {
d[e.to] = (d[e.from] - e.com) * e.rate;

if (i == v - 1) {
return true;
}
}
}
}
return false;
}

int main(int argc, char const *argv[]) {
int n, m, s;
double v;
scanf("%d%d%d%lf", &n, &m, &s, &v);
s--;
for (int i = 0; i < 2 * m; i += 2) {
int foo, bar;
double r1, c1, r2, c2;
scanf("%d%d%lf%lf%lf%lf", &foo, &bar,
&es[i].rate, &es[i].com, &es[i+1].rate, &es[i+1].com);
foo--, bar--;
es[i].from = foo;
es[i].to = bar;
es[i+1].from = bar;
es[i+1].to = foo;
}

puts(find_negative_loop(s, n, 2 * m, v) ? "YES" : "NO");

return 0;
}