Submission #847591

#	Time	Username	Problem	Language	Result	Execution time	Memory
847591		math_rabbit_1028	봉쇄 시간 (IOI23_closing)	C++17	35 / 100	131 ms	158684 KiB

closing

#include "closing.h"
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;

int n, x, y, ch[202020], ans;
ll k, need[202020], d[202020], dsum[202020];
vector<pll> adj[202020];

pair<int, ll> in[3030][3030];
void inner() {
    for (int r = x; r <= y; r++) {
        in[r][y].first = r - x + 2;
        in[r][y].second = 0;
        ll need[3030];
        for (int i = x; i <= r; i++) need[i] = dsum[i] - dsum[x];
        for (int i = r + 1; i <= y; i++) need[i] = 0;
        for (int i = x; i <= y; i++) in[r][y].second += need[i];
        for (int l = y - 1; l >= x; l--) {
            in[r][l].first = in[r][l + 1].first + 1;
            in[r][l].second = in[r][l + 1].second + max(0LL, dsum[y] - dsum[l] - need[l]);
        }
    }
}

ll dp[404040], nxt[404040];
void update(ll a, ll b) {
    for (int i = 0; i <= 2*n; i++) {
        ll val = dp[i];
        if (i >= 1) val = min(val, dp[i - 1] + a);
        if (i >= 2) val = min(val, dp[i - 2] + b);
        nxt[i] = val;
    }
    for (int i = 0; i <= 2*n; i++) dp[i] = nxt[i];
}
void outer() {
    for (int i = 0; i <= 2*n; i++) dp[i] = 2e18;
    dp[0] = 0;

    for (int i = 0; i < x; i++) {
        update(dsum[x] - dsum[i], dsum[y] - dsum[i]);
    }
    for (int i = y + 1; i < n; i++) {
        update(dsum[i] - dsum[y], dsum[i] - dsum[x]);
    }
}

int max_score(int N, int X, int Y, ll K, vector<int> U, vector<int> V, vector<int> W) {
    n = N, x = X, y = Y, k = K;

    ans = 0;
    for (int i = 0; i < n; i++) adj[i].clear();

    for (int i = 0; i < (int)U.size(); i++) {
        adj[U[i]].push_back({V[i], W[i]});
        adj[V[i]].push_back({U[i], W[i]});
        d[max(U[i], V[i])] = W[i];
    }

    for (int i = 1; i < n; i++) dsum[i] = dsum[i - 1] + d[i];

    inner();
    outer();

    for (int r = x; r <= y; r++) {
        for (int l = y; l >= x; l--) {
            if (k - in[r][l].second < 0) continue;
            int val = in[r][l].first;
            int idx = upper_bound(dp, dp + 2*n + 1, k - in[r][l].second) - dp - 1;
            ans = max(ans, val + idx);
        }
    }

    return ans;
}

• Subtask 0: 0 / 0
• Subtask 1: 0 / 8
• Subtask 2: 9 / 9
• Subtask 3: 12 / 12
• Subtask 4: 14 / 14
• Subtask 5: 0 / 9
• Subtask 6: 0 / 11
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 17