possivel solução da lista

#include <iostream>
#include <cstring>
#include <limits>

using namespace std;

const int INF = numeric_limits<int>::max();
const int MAX_Q = 300;

int graph[MAX_Q][MAX_Q];
int dist[MAX_Q];
bool visited[MAX_Q];

void dijkstra(int start, int Q) {
    for (int i = 0; i < Q; ++i) {
        dist[i] = INF;
    }
    memset(visited, false, sizeof(visited));
    dist[start] = 0;

    for (int i = 0; i < Q; ++i) {
        int u = -1;
        for (int j = 0; j < Q; ++j) {
            if (!visited[j] && (u == -1 || dist[j] < dist[u])) {
                u = j;
            }
        }

        if (dist[u] == INF) break;

        visited[u] = true;

        for (int v = 0; v < Q; ++v) {
            if (graph[u][v] != INF && dist[u] + graph[u][v] < dist[v]) {
                dist[v] = dist[u] + graph[u][v];
            }
        }
    }
}

int main() {
    int Q, R, N;
    cin >> Q >> R >> N;

    for (int i = 0; i < Q; ++i) {
        for (int j = 0; j < Q; ++j) {
            graph[i][j] = (i == j) ? 0 : INF;
        }
    }

    for (int i = 0; i < R; ++i) {
        int A, B, M;
        cin >> A >> B >> M;
        graph[A][B] = M;
    }

    for (int i = 0; i < N; ++i) {
        int type;
        cin >> type;

        if (type == 1) {
            int A, B, M;
            cin >> A >> B >> M;
            graph[A][B] = M;
        } else if (type == 2) {
            int A, B;
            cin >> A >> B;

            dijkstra(A, Q);

            if (dist[B] == INF) {
                cout << -1 << endl;
            } else {
                cout << dist[B] << endl;
            }
        }
    }

    return 0;
}