Frequent values poj3368 线段树

Description

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You are given a sequence of n integers a1 , a2 , … , an in non-decreasing order. In addition to that, you are given several queries consisting of indices i and j (1 ≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integers ai , … , aj.

Input

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The input consists of several test cases. Each test case starts with a line containing two integers n and q (1 ≤ n, q ≤ 100000). The next line contains n integers a1 , … , an (-100000 ≤ ai ≤ 100000, for each i ∈ {1, …, n}) separated by spaces. You can assume that for each i ∈ {1, …, n-1}: ai ≤ ai+1. The following q lines contain one query each, consisting of two integers i and j (1 ≤ i ≤ j ≤ n), which indicate the boundary indices for the
query.

The last test case is followed by a line containing a single 0.

Output

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For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.

Source

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Ulm Local 2007

Solution

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题目给定的条件是数列不下降，那么可以线段树记录一段区间左起连续最长相同，右起连续最长相同，最长连续相同。当然rmq也是可以的。

Code

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#include <stdio.h>#include <string.h>#define rep(i, st, ed) for (int i = st; i <= ed; i += 1)#define fill(x, t) memset(x, t, sizeof(x))#define max(x, y) ((x)>(y)?(x):(y))#define min(x, y) ((x)<(y)?(x):(y))#define N 500005int ls[N], rs[N], mx[N], a[N];int read() {    char ch = getchar(); int x = 0, v = 1;    for (; ch < '0' || ch > '9'; v = ((ch == '-')?(-1):(v)), ch = getchar());    for (; ch <= '9' && ch >= '0'; x = x * 10 + ch - '0', ch = getchar());    return x * v;}int query(int now, int tl, int tr, int l, int r) {    if (tl == l && tr == r) {        return mx[now];    } else {        int mid = (tl + tr) >> 1;        if (r <= mid) {            return query(now << 1, tl, mid, l, r);        } else if (l > mid) {            return query(now << 1 | 1, mid + 1, tr, l, r);        } else {            int ret = max(query(now << 1, tl, mid, l, mid), query(now << 1 | 1, mid + 1, tr, mid + 1, r));            if (a[mid] == a[mid + 1]) {                ret = max(ret, min(mid - l + 1, rs[now << 1]) + min(r - mid, ls[now << 1 | 1]));            }            return ret;        }    }}void buildTree(int now, int l, int r) {    if (l == r) {        a[l] = read();        rs[now] = ls[now] = mx[now] = 1;        return ;    }    int mid = (l + r) >> 1;    buildTree(now << 1, l, mid);    buildTree(now << 1 | 1, mid + 1, r);    ls[now] = ls[now << 1];    rs[now] = rs[now << 1 | 1];    mx[now] = 1;    if (a[mid] == a[mid + 1]) {        if (a[l] == a[mid + 1]) {            ls[now] += ls[now << 1 | 1];        }        if (a[r] == a[mid]) {            rs[now] += rs[now << 1];        }        mx[now] = rs[now << 1] + ls[now << 1 | 1];    }    mx[now] = max(mx[now << 1], max(mx[now << 1 | 1], mx[now]));}int main(void) {    for (; true; ){        int n = read();        if (!n) {            return 0;        }        int q = read();        buildTree(1, 1, n);        while (q --) {            int tl = read();            int tr = read();            int ans = query(1, 1, n, tl, tr);            printf("%d\n", ans);        }    }    return 0;}