Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P3199 [HNOI2009]最小圈

P3199 [HNOI2009]最小圈

分数规划 + 负环 套路题。

一个 trick spfa 用于判负环时用栈,最短路时用队列。

查看代码
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#include <iostream>
#include <cstdio>
#include <queue>
using namespace std;
const double eps = 1e-10, inf = 1e7;
const int N = 3e3 + 10, M = 1e4 + 10;
bool vis[N];
int n, m, cnt[N];
double d[N], wt[M];
int idx = -1, hd[N], nxt[M], edg[M];
bool check(double mid)
{
for (int i = 1; i <= n; i++)
d[i] = cnt[i] = 0;
queue<int> q;
for (int i = 1; i <= n; i++)
{
q.push(i);
vis[i] = true;
}
while (!q.empty())
{
int t = q.front();
q.pop();
vis[t] = false;
for (int i = hd[t]; ~i; i = nxt[i])
if (d[t] + wt[i] - mid < d[edg[i]])
{
d[edg[i]] = d[t] + wt[i] - mid;
cnt[edg[i]] = cnt[t] + 1;
if (cnt[edg[i]] >= n)
return true;
if (!vis[edg[i]])
{
q.push(edg[i]);
vis[edg[i]] = true;
}
}
}
return false;
}
void add(int x, int y, double z)
{
nxt[++idx] = hd[x];
hd[x] = idx;
edg[idx] = y;
wt[idx] = z;
}
int main()
{
cin >> n >> m;
for (int i = 1; i <= n; i++)
hd[i] = -1;
for (int i = 1, a, b; i <= m; i++)
{
double c;
cin >> a >> b >> c;
add(a, b, c);
}
double l = -inf, r = inf, mid;
while (r - l > eps)
{
mid = (l + r) / 2;
if (check(mid))
r = mid;
else
l = mid;
}
printf("%.8lf", l);
return 0;
}