CSES - Aalto Competitive Programming 2024 - wk7 Homework - Results
Submission details
Task:Download Speed
Sender:louaha1
Submission time:2024-10-24 09:37:36 +0300
Language:C++ (C++11)
Status:READY
Result:
Test results
testverdicttime
#1ACCEPTED0.00 sdetails
#2ACCEPTED0.00 sdetails
#3ACCEPTED0.00 sdetails
#40.00 sdetails
#50.01 sdetails
#6ACCEPTED0.01 sdetails
#7ACCEPTED0.01 sdetails
#8ACCEPTED0.00 sdetails
#90.00 sdetails
#10ACCEPTED0.00 sdetails
#110.01 sdetails
#12ACCEPTED0.00 sdetails

Code

#include <iostream>
#include <vector>
#include <queue>
#include <climits>
#include <cstring>

using namespace std;

const int MAXN = 505; // Maximum number of nodes
int capacity[MAXN][MAXN]; // Capacity matrix
vector<int> adj[MAXN]; // Adjacency list

// Perform BFS to find an augmenting path, and record the path in 'parent'
int bfs(int s, int t, vector<int>& parent) {
    fill(parent.begin(), parent.end(), -1);
    parent[s] = s;
    queue<pair<int, int>> q;
    q.push({s, INT_MAX});

    while (!q.empty()) {
        int u = q.front().first;
        int flow = q.front().second;
        q.pop();

        // Traverse all neighbors of u
        for (int v : adj[u]) {
            // If the node is not yet visited and the capacity is greater than 0
            if (parent[v] == -1 && capacity[u][v] > 0) {
                parent[v] = u;
                int new_flow = min(flow, capacity[u][v]);

                // If we reached the sink, return the flow
                if (v == t)
                    return new_flow;

                q.push({v, new_flow});
            }
        }
    }

    return 0; // No path found
}

// Edmonds-Karp algorithm to compute the maximum flow
int edmonds_karp(int s, int t, int n) {
    int flow = 0;
    vector<int> parent(n + 1);

    while (int new_flow = bfs(s, t, parent)) {
        flow += new_flow;

        // Update the capacities of the edges in the augmenting path
        int cur = t;
        while (cur != s) {
            int prev = parent[cur];
            capacity[prev][cur] -= new_flow;
            capacity[cur][prev] += new_flow;
            cur = prev;
        }
    }

    return flow;
}

int main() {
    int n, m;
    cin >> n >> m;

    // Initialize the capacity matrix and adjacency list
    memset(capacity, 0, sizeof(capacity));

    // Read the edges and build the graph
    for (int i = 0; i < m; i++) {
        int a, b, c;
        cin >> a >> b >> c;
        capacity[a][b] += c;
        capacity[b][a] += c; // Since the graph is undirected, both directions have the same capacity
        adj[a].push_back(b);
        adj[b].push_back(a);
    }

    // Run the Edmonds-Karp algorithm to find the maximum flow from node 1 to node n
    int result = edmonds_karp(1, n, n);
    cout << result << endl;

    return 0;
}

Test details

Test 1

Verdict: ACCEPTED

input
4 3
1 2 5
2 3 3
3 4 6

correct output
3

user output
3

Test 2

Verdict: ACCEPTED

input
4 5
1 2 1
1 3 1
2 3 1
2 4 1
...

correct output
2

user output
2

Test 3

Verdict: ACCEPTED

input
4 5
1 2 1000000000
1 3 1000000000
2 3 1
2 4 1000000000
...

correct output
2000000000

user output
2000000000

Test 4

Verdict:

input
2 1
2 1 100

correct output
0

user output
100

Test 5

Verdict:

input
2 1000
1 2 1000000000
1 2 1000000000
1 2 1000000000
1 2 1000000000
...

correct output
1000000000000

user output
0

Test 6

Verdict: ACCEPTED

input
500 998
1 2 54
1 3 59
1 4 83
2 5 79
...

correct output
60

user output
60

Test 7

Verdict: ACCEPTED

input
500 998
1 2 530873053
1 3 156306296
1 4 478040476
3 5 303609600
...

correct output
1093765123

user output
1093765123

Test 8

Verdict: ACCEPTED

input
2 1
1 2 1

correct output
1

user output
1

Test 9

Verdict:

input
4 5
1 2 3
2 4 2
1 3 4
3 4 5
...

correct output
6

user output
9

Test 10

Verdict: ACCEPTED

input
4 5
1 2 1
1 3 2
3 2 1
2 4 2
...

correct output
3

user output
3

Test 11

Verdict:

input
10 999
1 2 1000000000
1 2 1000000000
1 2 1000000000
1 2 1000000000
...

correct output
111000000000

user output
0

Test 12

Verdict: ACCEPTED

input
7 9
1 2 1
1 3 1
1 4 1
2 5 1
...

correct output
2

user output
2