Submission #758128

#	Time	Username	Problem	Language	Result	Execution time	Memory
758128		PixelCat	Rope (JOI17_rope)	C++14	55 / 100	1115 ms	262144 KiB

rope

#include <bits/stdc++.h>
#define For(i, a, b) for(int i = a; i <= b; i++)
#define Forr(i, a, b) for(int i = a; i >= b; i--)
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define sz(x) ((int)x.size())
#define eb emplace_back
#define int LL
#define INF (int)(1e18)
using namespace std;
using LL = long long;
using pii = pair<int, int>;

void chmin(int &x, int a) {
    x = min(x, a);
}
void chmax(int &x, int a) {
    x = max(x, a);
}

const int MAXN = 1000010;

int ans[MAXN];
int cnt1[MAXN];

void solve(int n, int m, vector<pii> v2, vector<int> v1) {
    for(auto &i:v2) {
        if(i.F > i.S) swap(i.F, i.S);
    }  // i.F <= i.S

    memset(cnt1, 0, sizeof(cnt1));
    map<pii, int> cnt2;
    for(auto &i:v2) {
        cnt1[i.F]++;
        cnt1[i.S]++;
        cnt2[i]++;
    }

    For(c1, 1, m) {
        For(c2, c1 + 1, m) {
            int cost = n - cnt1[c1] - cnt1[c2] + cnt2[pii(c1, c2)];
            for(auto &i:v1) {
                cost -= max(i == c1, i == c2);
            }
            chmin(ans[c1], cost);
            chmin(ans[c2], cost);
        }
        int cost = n - cnt1[c1];
        for(auto &i:v1) cost -= (i == c1);
        chmin(ans[c1], cost);
    }
}

int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    // =^-w-^=
    int n, m; cin >> n >> m;
    vector<int> a(n);
    for(auto &i:a) cin >> i;

    For(i, 1, m) {
        ans[i] = INF;
    }

    if(n % 2) {
        For(_, 0, 1) {
            vector<pii> v2;
            for(int i = 1; i < n; i += 2) {
                v2.eb(pii(a[i - 1], a[i]));
            }
            solve(n, m, v2, { a.back() });
            reverse(all(a));
        }
    } else {
        vector<pii> v2;
        for(int i = 1; i < n; i += 2) {
            v2.eb(pii(a[i - 1], a[i]));
        }
        solve(n, m, v2, {});

        v2.clear();
        for(int i = 2; i < n; i += 2) {
            v2.eb(pii(a[i - 1], a[i]));
        }
        solve(n, m, v2, { a.front(), a.back() });
    }

    For(i, 1, m) cout << ans[i] << "\n";
    return 0;
}

• Subtask 1: 15 / 15
• Subtask 2: 30 / 30
• Subtask 3: 10 / 10
• Subtask 4: 0 / 25
• Subtask 5: 0 / 20