Submission #233606

#	Time	Username	Problem	Language	Result	Execution time	Memory
233606		dolphingarlic	Islands (IOI08_islands)	C++14	80 / 100	636 ms	131076 KiB

islands

/*
IOI 2008 Islands
- First, split the graph up into its connected components. We want the sum of the longest
  paths in each component
- Notice how each component is made of a single cycle with trees appended to each node on the cycle
- Case 1: The longest path lies in one of the trees
    - Just use DP to find this length
- Case 2: The longest path goes from one tree to the other and crosses the cycle
    - Let dp[i] = The length of the longest path going through node i
    - If we flatten the cycle by removing an edge, then notice how
      we can use prefix sums to get the answer in this case
- Just handle these 2 cases for each component
- Complexity: O(N)
*/

#include <bits/stdc++.h>
#define FOR(i, x, y) for (int i = x; i < y; i++)
typedef long long ll;
using namespace std;

struct Edge {
    int idx, first;
    ll second;

    bool operator!=(Edge B) { return idx != B.idx; }
};

vector<Edge> graph[1010101], cycle;
ll discarded, ans = 0, cmp_ans, dp[1010101], p[5][1010101];
int cycle_start = 0;
bool processed[1010101], visited[1010101], on_cycle[1010101], adding = false;

void find_cycle(int node, Edge parent = {0, 0, 0}) {
    if (visited[node]) {
        cycle_start = node;
        cycle.push_back({0, node, 0});
        cycle.push_back(parent);
        adding = true;
        return;
    }
    visited[node] = true;
    for (Edge i : graph[node]) if (i != parent) {
        find_cycle(i.first, {i.idx, node, i.second});
        if (cycle_start) {
            if (node == cycle_start) adding = false;
            if (adding) cycle.push_back(parent);
            return;
        }
    }
}

void calc_tree(int node, int parent = 0) {
    processed[node] = true;
    ll mx = 0, sc = 0;
    for (Edge i : graph[node]) if (i.first != parent && !on_cycle[i.first]) {
        calc_tree(i.first, node);
        if (dp[i.first] + i.second > mx) sc = mx, mx = dp[i.first] + i.second;
        else if (dp[i.first] + i.second > sc) sc = dp[i.first] + i.second;
    }
    dp[node] = mx;
    cmp_ans = max(cmp_ans, mx + sc);
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    FOR(i, 1, n + 1) {
        int v;
        ll l;
        cin >> v >> l;
        graph[i].push_back({i, v, l});
        graph[v].push_back({i, i, l});
    }

    FOR(i, 1, n + 1) if (!processed[i]) {
        cycle.clear();
        cycle_start = 0;
        find_cycle(i);
        discarded = cycle.back().second;
        cycle.pop_back();

        for (Edge j : cycle) on_cycle[j.first] = true;

        cmp_ans = 0;
        int m = cycle.size();

        FOR(j, 1, m) p[0][j] = p[0][j - 1] + cycle[j].second;
        for (int j = m - 2; j >= 0; j--) p[1][j] = p[1][j + 1] + cycle[j + 1].second;

        FOR(j, 0, m) {
            calc_tree(cycle[j].first);
            p[2][j] = dp[cycle[j].first] - p[0][j];
            p[3][j] = dp[cycle[j].first] + p[0][j];
            p[4][j] = dp[cycle[j].first] + p[1][j];
        }

        ll mx = 0;
        FOR(j, 0, m) {
            cmp_ans = max(cmp_ans, p[3][j] + mx);
            mx = max(mx, p[2][j]);
        }
        mx = p[4][m - 1];
        for (int j = m - 2; j >= 0; j--) {
            cmp_ans = max(cmp_ans, p[3][j] + mx + discarded);
            mx = max(mx, p[4][j]);
        }

        ans += cmp_ans;

        FOR(j, 0, 5) FOR(k, 0, m) p[j][k] = 0;
    }
    cout << ans;
    return 0;
}

• Subtask 1: 33 / 33
• Subtask 2: 3 / 3
• Subtask 3: 4 / 4
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10