71 lines
1.8 KiB
C++
71 lines
1.8 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const double eps = 1e-8;
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const int INF = 0x3f3f3f3f;
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const int N = 1010, M = 5010;
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int n, m;
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int f[N], cnt[N];
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double dist[N];
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bool st[N];
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// 邻接表
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int idx, h[N], e[M], w[M], ne[M];
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void add(int a, int b, int c) {
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e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
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}
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bool check(double mid) {
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queue<int> q;
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memset(cnt, 0, sizeof cnt);
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memset(dist, 0x3f, sizeof dist);
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memset(st, false, sizeof st);
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for (int i = 1; i <= n; i++) {
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q.push(i);
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st[i] = 1;
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}
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while (q.size()) {
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int u = q.front();
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q.pop();
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st[u] = 0;
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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// 最短路
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if (dist[v] > dist[u] + w[i] * mid - f[u]) {
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dist[v] = dist[u] + w[i] * mid - f[u];
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// 判负环
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cnt[v] = cnt[u] + 1;
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if (cnt[v] >= n) return 1;
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if (!st[v]) {
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q.push(v);
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st[v] = 1;
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}
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}
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}
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}
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return 0;
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}
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int main() {
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scanf("%d %d", &n, &m);
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for (int i = 1; i <= n; i++) scanf("%d", &f[i]); // 每个点都有一个权值f[i]
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// 初始化邻接表
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memset(h, -1, sizeof h);
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int a, b, c;
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while (m--) {
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scanf("%d %d %d", &a, &b, &c);
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add(a, b, c);
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}
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// 浮点数二分
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double l = 0, r = INF;
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while (r - l > eps) {
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double mid = (l + r) / 2;
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if (check(mid))
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l = mid; // 存在负环时，mid再大一点,最终取得01分数规则的最大值
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else
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r = mid; // 不存在负环时，mid再小一点
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}
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printf("%.2lf\n", l);
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return 0;
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} |