uva 11235 - Frequent values(RMQ)

题目链接：uva 11235 - Frequent values

题目大意：给定一个非降序的整数数组，要求计算对于一些询问(i,j)，回答ai,ai+1,…,aj中出现最多的数出现的次数。

解题思路：因为序列为非降序的，所以相同的数字肯定是靠在一起的，所以用o(n)的方法处理处每段相同数字的区间。然后对于每次询问：

• num[i]=num[j]:j−i+1
• numi≠numj:max(RMQ(righti+1,reftj−1),max(righti−i+1,j−leftj+1))

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 1e5+5;int N, Q, num[maxn], rmq[maxn][20];int left[maxn], right[maxn];void RMQ_init () {    memset(rmq, 0, sizeof(rmq));    for (int i = 1; i <= N; i++)        rmq[i][0] = right[i] - left[i] + 1;    for (int j = 1; (1<<j) <= N; j++) {        for (int i = 1; i + (1<<j) - 1 <= N; i++)            rmq[i][j] = max(rmq[i][j-1], rmq[i+(1<<(j-1))][j-1]);    }}void init () {    for (int i = 1; i <= N; i++)        scanf("%d", &num[i]);    left[1] = 1;    for (int i = 2; i <= N; i++) {        if (num[i] == num[i-1])            left[i] = left[i-1];        else            left[i] = i;    }    right[N] = N;    for (int i = N-1; i >= 1; i--) {        if (num[i] == num[i+1])            right[i] = right[i+1];        else            right[i] = i;    }    RMQ_init();}int RMQ (int L, int R) {    if (L > R)        return 0;    int k = 0;    while (1<<(k+1) <= R-L+1)        k++;    return max(rmq[L][k], rmq[R-(1<<k)+1][k]);}int main () {    while (scanf("%d%d", &N, &Q) == 2 && N) {        init();        int x, y;        for (int i = 0; i < Q; i++) {            scanf("%d%d", &x, &y);            if (num[x] == num[y])                printf("%d\n", y - x + 1);            else                printf("%d\n", max(RMQ(right[x]+1, left[y]-1), max(right[x] - x + 1, y - left[y] + 1)));        }    }    return 0;}