子图最短路(Subgraph Shortest Paths) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m;
    if (!(cin >> n >> m)) return 0;
    const long long INF = (1LL << 60);
    const long long MOD = 1000000000LL;
    vector<vector<long long>> adj(n + 1, vector<long long>(n + 1, INF));
    for (int i = 1; i <= n; ++i) adj[i][i] = 0;
    for (int i = 0; i < m; ++i) {
        int u, v;
        long long w;
        cin >> u >> v >> w;
        adj[u][v] = min(adj[u][v], w);
        adj[v][u] = min(adj[v][u], w);
    }
    long long ans = 0;
    vector<vector<long long>> dist(n + 1, vector<long long>(n + 1, INF));
    for (int l = 1; l <= n; ++l) {
        for (int i = l; i <= n; ++i) {
            for (int j = l; j <= n; ++j) {
                dist[i][j] = (i == j ? 0 : INF);
            }
        }
        for (int r = l; r <= n; ++r) {
            for (int i = l; i < r; ++i) {
                long long best = adj[i][r];
                for (int j = l; j < r; ++j) {
                    if (dist[i][j] < INF && adj[j][r] < INF) {
                        best = min(best, dist[i][j] + adj[j][r]);
                    }
                }
                dist[i][r] = dist[r][i] = best;
            }
            for (int i = l; i < r; ++i) {
                if (dist[i][r] >= INF) continue;
                for (int j = i + 1; j < r; ++j) {
                    if (dist[r][j] < INF && dist[i][r] + dist[r][j] < dist[i][j]) {
                        dist[i][j] = dist[j][i] = dist[i][r] + dist[r][j];
                    }
                }
            }
            long long sub_sum = 0;
            for (int i = l; i <= r; ++i) {
                for (int j = i + 1; j <= r; ++j) {
                    if (dist[i][j] < INF) {
                        sub_sum += dist[i][j];
                    }
                }
            }
            ans = (ans + (sub_sum % MOD)) % MOD;
        }
    }
    cout << ans << "\n";
    return 0;
}