CSES - Aalto Competitive Programming 2024 - wk7 Homework - Results
Submission details
Task:Download Speed
Sender:bubu2006
Submission time:2024-10-10 00:07:20 +0300
Language:C++ (C++20)
Status:READY
Result:ACCEPTED
Test results
testverdicttime
#1ACCEPTED0.00 sdetails
#2ACCEPTED0.00 sdetails
#3ACCEPTED0.00 sdetails
#4ACCEPTED0.00 sdetails
#5ACCEPTED0.00 sdetails
#6ACCEPTED0.00 sdetails
#7ACCEPTED0.01 sdetails
#8ACCEPTED0.00 sdetails
#9ACCEPTED0.00 sdetails
#10ACCEPTED0.00 sdetails
#11ACCEPTED0.00 sdetails
#12ACCEPTED0.00 sdetails

Code

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

#define int long long
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

/**
 * Author: chilli
 * Date: 2019-04-26
 * License: CC0
 * Source: https://cp-algorithms.com/graph/dinic.html
 * Description: Flow algorithm with complexity $O(VE\log U)$ where $U = \max |\text{cap}|$.
 * $O(\min(E^{1/2}, V^{2/3})E)$ if $U = 1$; $O(\sqrt{V}E)$ for bipartite matching.
 * Status: Tested on SPOJ FASTFLOW and SPOJ MATCHING, stress-tested
 */

struct Dinic {
    struct Edge {
        int to, rev;
        ll c, oc;
        ll flow() { return max(oc - c, 0LL); } // if you need flows
    };
    vi lvl, ptr, q;
    vector<vector<Edge>> adj;
    Dinic(int n) : lvl(n), ptr(n), q(n), adj(n) {}
    void addEdge(int a, int b, ll c, ll rcap = 0) {
        adj[a].push_back({b, sz(adj[b]), c, c});
        adj[b].push_back({a, sz(adj[a]) - 1, rcap, rcap});
    }
    ll dfs(int v, int t, ll f) {
        if (v == t || !f) return f;
        for (int& i = ptr[v]; i < sz(adj[v]); i++) {
            Edge& e = adj[v][i];
            if (lvl[e.to] == lvl[v] + 1)
                if (ll p = dfs(e.to, t, min(f, e.c))) {
                    e.c -= p, adj[e.to][e.rev].c += p;
                    return p;
                }
        }
        return 0;
    }
    ll calc(int s, int t) {
        ll flow = 0; q[0] = s;
        rep(L,0,31) do { // 'int L=30' maybe faster for random data
            lvl = ptr = vi(sz(q));
            int qi = 0, qe = lvl[s] = 1;
            while (qi < qe && !lvl[t]) {
                int v = q[qi++];
                for (Edge e : adj[v])
                    if (!lvl[e.to] && e.c >> (30 - L))
                        q[qe++] = e.to, lvl[e.to] = lvl[v] + 1;
            }
            while (ll p = dfs(s, t, LLONG_MAX)) flow += p;
        } while (lvl[t]);
        return flow;
    }
    bool leftOfMinCut(int a) { return lvl[a] != 0; }
};

signed main() {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit); // RTE if input wrong datatype

    int n, m;
    cin >> n >> m;

    Dinic dinic(n);
    rep(i, 0, m) {
        int a, b, c;
        cin >> a >> b >> c;
        a--; b--;
        dinic.addEdge(a, b, c);
    }
    cout << dinic.calc(0, n - 1) << '\n';
}

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: ACCEPTED

input
2 1
2 1 100

correct output
0

user output
0

Test 5

Verdict: ACCEPTED

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

correct output
1000000000000

user output
1000000000000

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: ACCEPTED

input
4 5
1 2 3
2 4 2
1 3 4
3 4 5
...

correct output
6

user output
6

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: ACCEPTED

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

correct output
111000000000

user output
111000000000

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