2023icpc南京站补题

思路简记

建图后，可以枚举每个点作为起点跑 Dijkstra。如果最短路过程中再次回到起点，就得到经过该点的一个最小环候选值；再结合题目给定的阈值判断答案。

代码

#include <bits/stdc++.h>
using namespace std;

const int N = 2010;
const int INF = 0x3f3f3f3f;

vector<pair<int, int>> e[N];
int n, m, c, minn, dist[N];
bool st[N];

void Dijkstra(int u) {
    minn = INF;
    for (int i = 1; i <= n; i++) {
        dist[i] = INF;
        st[i] = false;
    }

    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
    dist[u] = 0;
    q.push({dist[u], u});

    while (!q.empty()) {
        auto [dis, ver] = q.top();
        q.pop();
        if (st[ver]) continue;
        st[ver] = true;

        for (auto [v, w] : e[ver]) {
            if (v == u) minn = min(minn, dis + w);
            if (st[v]) continue;
            if (dist[v] > dis + w) {
                dist[v] = dis + w;
                q.push({dist[v], v});
            }
        }
    }
}

void solve() {
    cin >> n >> m >> c;
    int ans = 0;

    for (int i = 1; i <= m; i++) {
        int u, v, z;
        cin >> u >> v >> z;
        e[u].push_back({v, z});
        if (z <= c) ans = 1;
    }

    for (int i = 1; i <= n; i++) {
        Dijkstra(i);
        if (minn <= c) {
            ans = 2;
            break;
        }
    }

    cout << ans << "\n";
}