【LOJ10121】与众不同

【LOJ10121】与众不同

题面

LOJ

题解

这题是_\(tham\)给\(ztl\)他们做的，然而这道题™居然还想了蛮久。。。

首先可以尺取出一个位置\(i\)上一个合法的最远位置\(pre_i\)

而对于一个询问\((l,r)\)，因为\(pre_i\)是单调的

所以可以二分出\(pre_i\geq l\)的第一个位置\(mid\)

用\(st\)表维护一下区间\(i-pre_i+1\)最大值\(qmax\)

则\(ans=max(mid-l,qmax(mid,r))\)

注意判断一下边界情况

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
inline int gi() {
    register int data = 0, w = 1;
    register char ch = 0;
    while (!isdigit(ch) && ch != '-') ch = getchar();
    if (ch == '-') w = -1, ch = getchar();
    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();
    return w * data;
}
const int MAX_N = 200005;
const int MAX_LOG_N = 19;
const int MAX_V = 1e6;
int bln[MAX_V << 1 | 1];
int N, M, a[MAX_N], pre[MAX_N];
int st[MAX_N][MAX_LOG_N], lg2[MAX_N];
void Prepare() {
    int l = 1, r = 0;
    do {
        bln[a[++r]]++;
        while (bln[a[r]] > 1) --bln[a[l++]];
        pre[r] = l;
    } while (l <= N && r <= N && l <= r);
    for (int i = 1; i <= N; i++) st[i][0] = i - pre[i] + 1;
    for (int j = 1; j <= 18; j++)
        for (int i = 1; i + (1 << j) - 1 <= N; i++)
            st[i][j] = max(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);
    for (int i = 2; i <= MAX_N; i++) lg2[i] = lg2[i >> 1] + 1;
}
int qmax(int l, int r) {
    int t = lg2[r - l + 1];
    return max(st[l][t], st[r - (1 << t) + 1][t]);
}
int solve(int ql, int qr) {
    if (qmax(qr, qr) >= qr - ql + 1) return qr - ql + 1;
    int l = ql, r = qr, res = qr;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (ql <= pre[mid]) res = mid, r = mid - 1;
        else l = mid + 1;
    }
    return max(res - ql, qmax(res, qr));
}
int main () {
    N = gi(), M = gi();
    for (int i = 1; i <= N; i++) a[i] = gi() + MAX_V;
    Prepare();
    while (M--) {
        int l = gi() + 1, r = gi() + 1;
        printf("%d\n", solve(l, r));
    }
    return 0;
}