负环 : 01分数规划

观光奶牛

\(求负环 + 01分数规划\)

思路

\(将点权算入边权:\)

\[\begin{aligned} &\frac{\sum{f_i}}{\sum{w_i}}>mid\\ \Rightarrow\ &\sum{f_i}-\sum{w_i}*mid>0\\ \Rightarrow\ &\sum{(f_i-mid*w_i)}>0 \end{aligned} \]

\(将每条边的边权转化成f_i-mid*w_i,于是问题转化成求正环.\)

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define pb push_back
#define PII pair<int, int>
#define x first
#define y second
#define inf 0x3f3f3f3f
const int N = 1010, M = 5010;
int n, m;
int wf[N];
int h[N], e[M], w[M], ne[M], idx;
double dist[N];
int q[N], cnt[N];
bool st[N];

void add(int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}

bool check(double mid) {
    memset(st, 0, sizeof st);
    memset(cnt, 0, sizeof cnt);
    int hh = 0, tt = 0;
    for (int i = 1; i <= n; ++i) {
        q[tt++] = i;
        st[i] = true;
    }
    while (hh != tt) {
        int t = q[hh++];
        if (hh == N) hh = 0;
        st[t] = false;
        for (int i = h[t]; ~i; i = ne[i]) {
            int j = e[i];
            if (dist[j] < dist[t] + wf[t] - mid * w[i]) {
                dist[j] = dist[t] + wf[t] - mid * w[i];
                cnt[j] += 1;
                if (cnt[j] >= n) return true;
                if (!st[j]) {
                    q[tt++] = j;
                    if (tt == N) tt = 0;
                    st[j] = true;
                }
            }
        }
    }
    return false;
}

int main() {
    IO;
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) cin >> wf[i];
    memset(h, -1, sizeof h);
    for (int j = 0; j < m; ++j) {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c);
    }
    double l = 0, r = 1e6;
    while (r - l > 1e-4) {
        double mid = (l + r) / 2;
        if (check(mid)) l = mid;
        else r = mid;
    }
    printf("%.2lf\n", l);
    return 0;
}