POJ3621或洛谷2868 [USACO07DEC]观光奶牛Sightseeing Cows

一道\(0/1\)分数规划+负环

POJ原题链接

洛谷原题链接

显然是\(0/1\)分数规划问题。

二分答案，设二分值为\(mid\)。

然后对二分进行判断，我们建立新图，没有点权，设当前有向边为\(z=(x,y)\)，\(time\)为原边权，\(fun\)为原点权，则将该边权换成\(mid\times time[z]+fun[x]\)，然后在上面跑\(SPFA\)。

如果有一个环使得\(\sum\{mid\times time[z]+fun[x]\}<0\)，则说明\(mid\)小了，而式子表示的意义就是原图存在负环；如果有环使得该式\(\geqslant0\)，那么说明答案不超过\(mid\)，这时\(SPFA\)就会正常结束。

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 1010;
const int M = 5010;
int fi[N], di[M], ne[M], da[M], fun[N], q[N << 10], cnt[N], l, n;
bool v[N];
double dis[N];
inline int re()
{
	int x = 0;
	char c = getchar();
	bool p = 0;
	for (; c<'0' || c>'9'; c = getchar())
		p |= c == '-';
	for (; c >= '0'&&c <= '9'; c = getchar())
		x = x * 10 + (c - '0');
	return p ? -x : x;
}
inline void add(int x, int y, int z)
{
	di[++l] = y;
	da[l] = z;
	ne[l] = fi[x];
	fi[x] = l;
}
bool judge(double mid)
{
	int head = 0, tail = 1, i, x, y;
	double z;
	bool p = 1;
	memset(dis, 66, sizeof(dis));
	memset(v, 0, sizeof(v));
	memset(cnt, 0, sizeof(cnt));
	dis[1] = 0;
	q[1] = 1;
	while (head != tail && p)
	{
		x = q[++head];
		v[x] = 0;
		for (i = fi[x]; i; i = ne[i])
		{
			y = di[i];
			z = mid * da[i] - fun[x];
			if (dis[y] > dis[x] + z)
			{
				dis[y] = dis[x] + z;
				cnt[y] = cnt[x] + 1;
				if (cnt[y] >= n)
				{
					p = 0;
					break;
				}
				if (!v[y])
				{
					q[++tail] = y;
					v[y] = 1;
				}
			}
		}
	}
	if (p)
		return false;
	return true;
}
int main()
{
	int i, m, x, y, z;
	double l = 0, r = 0, mid;
	n = re();
	m = re();
	for (i = 1; i <= n; i++)
		fun[i] = re();
	for (i = 1; i <= m; i++)
	{
		x = re();
		y = re();
		z = re();
		r += z;
		add(x, y, z);
	}
	r /= 2;
	while (l + 1e-3 < r)
	{
		mid = (l + r) / 2;
		if (judge(mid))
			l = mid;
		else
			r = mid;
	}
	printf("%.2f", l);
	return 0;
}