[POJ3368][UVA11235] Frequent values[ST表]

题意：给出一个不降序列，有多个询问，询问[l,r]中出现次数最多的数的出现次数

多组数据

对于序列-1 -1 1 1 1 1 3 10 10 10

可以这么理解<-1,2>, <1, 4>, ❤️,1>, <10,3>

cnt[i]记录这个数字的出现次数,lef[i]记录左端点,righ[i]记录右端点,belong[i]代表第i个数字属于哪一块

把块和数量丢进st表

然后对于区间[l,r]，答案就是 \(min(r, righ[l])-l+1\)，\(r-max(l, lef[r])+1\)，\(RMQ(belong[l]+1,belong[r]-1)\)中最大的一个

#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;

const int inf = 0x3f3f3f3f;
inline void chmax(int &x, int y) {if (x < y) x = y;}
inline void chmin(int &x, int y) {if (x > y) x = y;}
const int MAXN = 1e5+7;
int n, m, st[21][MAXN], a[MAXN], len[MAXN], belong[MAXN], lef[MAXN], righ[MAXN], tot, cnt[MAXN], Pow[21], logg[MAXN];
inline int read() {
	int c = getchar(), f = 1, x = 0;
	while(!isdigit(c)) (c=='-')&&(f=-f), c = getchar();
	while(isdigit(c)) x = x * 10 + c - '0', c = getchar();
	return x * f;
}
inline int RMQ(int l, int r) {
	int k = logg[r-l+1];
	return max(st[k][l], st[k][r-Pow[k]+1]);
}
int main(void) {
	for(int i = 2; i <= 100000; ++i) logg[i] = logg[i>>1] + 1;
	for(int i = 0; i <= 20; ++i) Pow[i] = (1<<i);
	while((n = read())) {
		tot = 0; memset(cnt, 0, n*4);
		m = read();
		for(int i = 1; i <= n; ++i) a[i] = read();	
		for(int i = 1; i <= n; ++i) {
			if (a[i] != a[i-1]) lef[i] = i, belong[i] = ++tot;
			else lef[i] = lef[i-1], belong[i] = tot;
			++cnt[tot];
		}
		for(int i = n; i >= 1; --i) {
			if (a[i] != a[i+1]) righ[i] = i;
			else righ[i] = righ[i+1];
		}
		for(int i = 1; i <= tot; ++i) st[0][i] = cnt[i];
//		for(int i = 1; i <= n; ++i) cout << lef[i] << " \n"[i==n];
//		for(int i = 1; i <= n; ++i) cout << righ[i]<< " \n"[i==n];
//		for(int i = 1; i <= n; ++i) cout << belong[i]<<" \n"[i==n];
//		for(int i = 1; i <= tot; ++i) cout << cnt[i] << " \n"[i==n];
		int mm = logg[n];
		for(int i = 1; i <= mm; ++i)
			for(int j = 1; j <= tot; ++j)
				st[i][j] = max(st[i-1][j], st[i-1][j+Pow[i-1]]);
		while(m--) {
			int l = read(), r = read();
			if (belong[l] == belong[r]) printf("%d\n", r-l+1);
			else {
				int ans = min(r, righ[l]) - l + 1;
				chmax(ans, r - max(l, lef[r]) + 1);
				if (belong[l] + 1 <= belong[r] - 1) chmax(ans, RMQ(belong[l]+1, belong[r] - 1));
				printf("%d\n", ans);
			}
		}
	}
	return 0;
}