UVA - 11235 - Frequent values （RMQ）
UVA - 11235
| Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
University of Ulm Local Contest
Problem F: Frequent values
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 Specification
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 Specification
For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.
Sample Input
10 3 -1 -1 1 1 1 1 3 10 10 10 2 3 1 10 5 10 0
Sample Output
1 4 3A naive algorithm may not run in time!
Source
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Data Structures and Libraries :: Data Structures with Our-Own Libraries :: Tree-related Data Structures
Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 3. Data Structures :: Maintaining Interval Data :: Examples
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Data Structures and Libraries :: Data Structures with Our-Own Libraries :: Tree-related Data Structures
思路：搞了我半天才搞出来，也是够啦，最后才发现是数组下标从一开始，我这代码凑合着用吧，因为是非降序的，所以进行游程编码（RLE），再算出区间内的最大值就好了
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100005;
int a[maxn];
int value[maxn], coun[maxn];
int num[maxn], le[maxn], ri[maxn];
int d[maxn][400];
int maxnum;//最大编号
void RMQ_init() {
for(int i = 0; i < maxnum+1; i++) d[i][0] = coun[i];
for(int j = 1; (1 << j) <= maxnum+1; j++)
for(int i = 0; i + (1<<j) - 1 < maxnum+1; i++)
d[i][j] = max(d[i][j-1], d[i + (1 << (j-1))][j-1]);
}
int RMQ(int l, int r) {
int k = 0;
while((1<<(k+1)) <= r - l + 1) k++;
return max(d[l][k], d[r-(1<<k) +1][k]);
}
int main() {
int n, q;
while(scanf("%d", &n) && n!=0) {
scanf("%d", &q);
for(int i = 0; i < n; i++)
scanf("%d", &a[i]);
maxnum = 0;
int cnt = 1;//当前出现次数
value[0] = a[0], coun[0] = cnt, num[0] = 0, le[0] = 0, ri[0] = 0;
for(int i = 1; i < n; i++) {
if(a[i] != a[i-1]) {
ri[maxnum] = i - 1;
cnt = 1; maxnum++;
le[maxnum] = i; ri[maxnum] = i;
num[i] = maxnum;
value[maxnum] = a[i];
coun[maxnum] = cnt;
}
else {
cnt++;
coun[maxnum] = cnt;
num[i] = maxnum;
ri[maxnum] = i;
}
}
RMQ_init();
for(int i = 0; i < q; i++) {
int l, r;
scanf("%d %d", &l, &r);
l--; r--;
//printf("%d %d %d\n", ri[num[l]]-l+1, r-le[num[r]]+1, RMQ(num[l]+1, num[r]-1));
if(num[l] == num[r]) printf("%d\n", r - l + 1);
else if(num[r] - num[l] == 1) printf("%d\n", max(ri[num[l]]-l+1, r-le[num[r]]+1));
else printf("%d\n", max(max(ri[num[l]]-l+1, r-le[num[r]]+1), RMQ(num[l]+1, num[r]-1)));
}
}
return 0;
}