题意

题目链接

给出长度为\(n\)的序列，每次询问区间\([l, r]\)，要求最大化

\(max |x ? y| : L_i ≤ x, y ≤ R_i and A_x = A_y\)

Sol

标算神仙的一批看不懂。

维护好每个数出现的左右位置之后直接上不删除莫队就行了

#include<bits/stdc++.h>

const int MAXN = 1e5 + 10, INF = 1e9 + 7;
using namespace std;
template<typename A, typename B> inline bool chmax(A &x, B y) {
    if(y > x) {x = y; return 1;}
    else return 0;
}
template<typename A, typename B> inline bool chmin(A &x, B y) {
    if(y < x) {x = y; return 1;}
    else return 0;
}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, K, M, a[MAXN], l[MAXN], r[MAXN], tl[MAXN], tr[MAXN], belong[MAXN], block, cnt, ans[MAXN];
struct query {
    int l, r, id;
    bool operator < (const query &rhs) const {
        return belong[l] == belong[rhs.l] ? r < rhs.r : belong[l] < belong[rhs.l];
    }
};
vector<query> q[MAXN];
int solve(int l, int r) {
    for(int i = l; i <= r; i++) tl[a[i]] = INF, tr[a[i]] = 0;
    int ans = 0;
    for(int i = l; i <= r; i++) {
        int x = a[i];
        chmin(tl[x], i), chmax(tr[x], i);
        chmax(ans, tr[x] - tl[x]);
    }
    return ans;
}
int update(int x) {
    chmax(tr[a[x]], x);
    chmin(tl[a[x]], x);
    return tr[a[x]] - tl[a[x]];
}
int update2(int x) {
    int v = a[x];
    chmax(r[v], x);
    chmin(l[v], x);
    return max(tr[v] - l[v], r[v] - l[v]);
}
void LxlDuLiu(vector<query> v, int id) {
    int base = id * block, ll = base, rr = ll - 1, pre = 0, now = 0;
    memset(tr, 0, sizeof(tr)); memset(r, 0, sizeof(r));
    memset(tl, 0x3f, sizeof(tl)); memset(l, 0x3f, sizeof(l));

    for(auto &x : v) {
        //memset(l, 0x3f, sizeof(l)); memset(r, 0, sizeof(r));
        while(rr < x.r) chmax(now, update(++rr));
        pre = now;
        while(ll > x.l) chmax(now, update2(--ll));
        chmax(ans[x.id], now);
        while(ll < base) r[a[ll]] = 0, l[a[ll]] = INF, ll++;
        now = pre;
    }
}
int main() {
//  freopen("a.in", "r", stdin);
    //freopen("b.out", "w", stdout);
    
    N = read(); K = read(); M = read(); block = sqrt(N); int mx = 0;
    for(int i = 1; i <= N; i++) a[i] = read(), belong[i] = (i - 1) / block + 1, chmax(mx, belong[i]);
    for(int i = 1; i <= M; i++) {
        int l = read(), r = read();
        if(belong[l] == belong[r]) ans[i] = solve(l, r);
        else q[belong[l]].push_back({l, r, i});
    }
    for(int i = 1; i <= mx; i++) sort(q[i].begin(), q[i].end());
    for(int i = 1; i <= mx; i++)
        LxlDuLiu(q[i], i);
    for(int i = 1; i <= M; i++) printf("%d\n", ans[i]);
    
    return 0;
}