Submission #1115458

#	Time	Username	Problem	Language	Result	Execution time	Memory
1115458		elush	Bitaro, who Leaps through Time (JOI19_timeleap)	C++14	34 / 100	3062 ms	47216 KiB

timeleap

#include <bits/stdc++.h>
#define x first
#define y second
#define all(v) v.begin(), v.end()
#define chkmax(a, b) a = max(a, b)
#define chkmin(a, b) a = min(a, b)
//#define int ll
using namespace std;
typedef long long ll;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef pair<ll, ll> pii;
typedef vector<pii> vii;

const int block = 550;
const int MAX_N = 3e5 + 5;

int nx[MAX_N][2];
pair<ll, int> dp[MAX_N][2];
pair<int, int> dq[2 * block + 1];
int rangesl[MAX_N / block + 1][block];
int rangesr[MAX_N / block + 1][block];
int ran[MAX_N][2];
int sz[MAX_N / block + 1];

void Update(int a, int b) {
    int dl = block, dr = block;
    for (int i = a; i < b; i++) {
        while (dl < dr) {
            if (ran[dq[dl].x][dq[dl].y] < ran[i][0]) {
                nx[dq[dl].x][dq[dl].y] = i;
                dl++;
            } else if (ran[dq[dr - 1].x][dq[dr - 1].y] > ran[i][1]) {
                nx[dq[dr - 1].x][dq[dr - 1].y] = i;
                dr--;
            } else break;
        }
        dq[--dl] = {i, 0}, dq[dr++] = {i, 1};
    }
    while (dl < dr) {
        nx[dq[dl].x][dq[dl].y] = b;
        dl++;
    }
    for (int i = b - 1; i >= a; i--) {
        for (int j = 0; j < 2; j++) {
            int x = nx[i][j];
            if (x == b) {
                dp[i][j] = {0, ran[i][j]};
            } else {
                if (ran[i][j] > ran[x][1]) {
                    dp[i][j] = {dp[x][1].x + ran[i][j] - ran[x][1], dp[x][1].y};
                } else {
                    dp[i][j] = {dp[x][0].x, dp[x][0].y};
                }
            }
        }
    }
    int bl = a / block;
    sz[bl] = b - a;
    rangesl[bl][0] = ran[a][0];
    rangesr[bl][0] = ran[a][1];
    for (int i = a + 1; i < b; i++) {
        rangesl[bl][i - a] = max(rangesl[bl][i - a - 1], ran[i][0]);
        rangesr[bl][i - a] = min(rangesr[bl][i - a - 1], ran[i][1]);
    }
}

ll Query(int a, int b, int c, int d) {
    ll ans = 0;
    while (a < c && a % block) {
        if (b < ran[a][0]) b = ran[a][0];
        else if (b > ran[a][1]) ans += b - ran[a][1], b = ran[a][1];
        a++;
    }
    while (a + block < c) {
        int bl = a / block;
        int i = min(lower_bound(rangesl[bl], rangesl[bl] + sz[bl], b) - rangesl[bl],
             lower_bound(rangesr[bl], rangesr[bl] + sz[bl], b, greater<int>()) - rangesr[bl]) + a;
        if (i < a + block) {
            if (b <= ran[i][0]) {
                ans += dp[i][0].x;
                b = dp[i][0].y;
            } else if (b >= ran[i][1]) {
                ans += dp[i][1].x;
                ans += b - ran[i][1];
                b = dp[i][1].y;
            }
        }
        a += block;
    }
    while (a < c) {
        if (b < ran[a][0]) b = ran[a][0];
        else if (b > ran[a][1]) ans += b - ran[a][1], b = ran[a][1];
        a++;
    }
    return ans + max<ll>(0, b - d);
}

int L[MAX_N], R[MAX_N], t[MAX_N], a[MAX_N], b[MAX_N], c[MAX_N], d[MAX_N];

ll ans[MAX_N];

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n, q;
    cin >> n >> q;
    for (int i = 0; i + 1 < n; i++) cin >> L[i] >> R[i];
    for (int i = 0; i < q; i++) {
        cin >> t[i] >> a[i] >> b[i] >> c[i];
        if (t[i] == 2) cin >> d[i];
    }
    for (int i = 0; i < q; i++) a[i]--, c[i]--;
    for (int i = 0; i + 1 < n; i++) R[i]--;
    for (int ti = 0; ti < 2; ti++) {
        for (int i = 0; i + 1 < n; i++) {
            ran[i][0] = L[i] - i, ran[i][1] = R[i] - i;
        }
        for (int i = 0; i + 1 < n; i += block) {
            Update(i, min(i + block, n - 1));
        }
        for (int i = 0; i < q; i++) {
            if (t[i] == 1) {
                ran[a[i]][0] = b[i] - a[i], ran[a[i]][1] = c[i] - a[i];
                Update(a[i] / block * block, min(n - 1, a[i] / block * block + block));
            } else {
                if (a[i] <= c[i]) ans[i] = Query(a[i], b[i] - a[i], c[i], d[i] - c[i]);
            }
        }
        reverse(L, L + n - 1), reverse(R, R + n - 1);
        for (int i = 0; i < q; i++) {
            if (t[i] == 1) a[i] = n - 2 - a[i];
            else a[i] = n - 1 - a[i], c[i] = n - 1 - c[i];
        }
    }
    for (int i = 0; i < q; i++) {
        if (t[i] == 2) cout << ans[i] << '\n';
    }
}

• Subtask 1: 4 / 4
• Subtask 2: 30 / 30
• Subtask 3: 0 / 66