题意
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("%dn", ans[i]);
return 0;
}内容来源于网络如有侵权请私信删除
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