【BZOJ5286】【HNOI2018】Turntable (Line Segment Tree)

Description

click me

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Solution

The first to find the answer is to$T_i$after the rotation$\max\{T_i-i\}+n-1$The minimum value of (consider waiting for a long time at the starting point, and then walk all the way without stopping)
Considering the conditions of the ring, it is very troublesome, so consider multiplying the sequence length, set$P_i=T_i-i$, then the answer becomes:$\displaystyle{\min_{i=1}^{n}\{\max_{j=1}^{n}\{P_{i+j-1}\}+i\}}+n-1$
find out if$j$enumerate to$2n-i+1$will not affect the answer.
Use the segment tree to maintain the interval in the interval$\max P$and$\min_{i=l}^{mid}\{i+\max_{i}^{r}P_i\}$That's it, in terms of maintenance, it is similar to the reconstruction of the BZOJ2957 building .

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Code

/************************************************
 * Au: Hany01
 * Date: Apr 23rd, 2018
 * Prob: [BZOJ5286][HNOI2018] 转盘
 * Email: [email protected]
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read()
{
    register int _, __; register char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 100005;

int mn[maxn << 4], mx[maxn << 4], n;

#define mid ((l + r) >> 1)
#define lc (t << 1)
#define rc (lc | 1)

int query(int t, int l, int r, int val)
{
    if (l == r) return l + max(val, mx[t]);
    if (val >= mx[rc]) return min(query(lc, l, mid, val), mid + 1 + val);
    return min(mn[t], query(rc, mid + 1, r, val));
}

void update(int t, int l, int r, int x, int dt)
{
    if (l == r) { mx[t] = dt; return ; }
    if (x <= mid) update(lc, l, mid, x, dt);
    else update(rc, mid + 1, r, x, dt);
    mx[t] = max(mx[lc], mx[rc]);
    if (l <= n) mn[t] = query(lc, l, mid, mx[rc]);
}

int main()
{
#ifdef hany01
    File("bzoj5286");
#endif

    static int m, p, x, y, las;

    n = read(), m = read(), p = read();
    For(i, 1, n)
        x = read(), update(1, 1, n << 1, i, x - i), update(1, 1, n << 1, i + n, x - i - n);
    printf("%d\n", las = (mn[1] + n - 1));

    while (m --)
        x = read() ^ (las * p), y = read() ^ (las * p),
        update(1, 1, n << 1, x, y - x), update(1, 1, n << 1, n + x, y - x - n),
        printf("%d\n", las = (mn[1] + n - 1));

    return 0;
}
//墙角数枝梅，凌寒独自开。
//    -- 王安石《梅》