hdu 1806 Frequent values（二分+线段树）

Frequent values

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1050 Accepted Submission(s): 375



[align=left]Problem Description[/align]
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 .

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.

[align=left]Sample Input[/align]

10 3
-1 -1 1 1 1 1 3 10 10 10
2 3
1 10
5 10
0

[align=left]Sample Output[/align]

1
4
3

样例：


其余你应该知道怎么做了！
[code]#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cstdlib>
using namespace std;

const int maxn = 100010;
int v[maxn] , sum[maxn];
struct tree{
int l , r , Max;
tree(int a = 0 , int b = 0 , int c = 0){
l = a , r = b , Max = c;
}
}a[4*maxn];
int n , q , cnt;

void build(int l , int r , int k){
a[k].l = l;
a[k].r = r;
if(l == r){
a[k].Max = v[l];
}else{
int mid = (l+r)/2;
build(l , mid , 2*k);
build(mid+1 , r , 2*k+1);
a[k].Max = max(a[2*k].Max , a[2*k+1].Max);
}
}

int querry(int l , int r , int k){
if(l <= a[k].l && a[k].r <= r){
return a[k].Max;
}else{
int mid = (a[k].l+a[k].r)/2;
if(mid >= r) return querry(l , r , 2*k);
else if(mid < l) return querry(l , r , 2*k+1);
else return max(querry(l , mid , 2*k) , querry(mid+1 , r , 2*k+1));
}
}

void initial(){
cnt = 1;
for(int i = 0; i < maxn; i++){
v[i] = 0;
sum[i] = 0;
}
}

void readcase(){
cin >> q;
int num , pre;
cin >> pre;
v[1] = 1;
for(int i = 1; i < n; i++){
scanf("%d" , &num);
if(num == pre) v[cnt]++;
else{
cnt++;
v[cnt]++;
pre = num;
}
}
for(int i = 1; i <= cnt; i++){
sum[i] = sum[i-1]+v[i];
}
}

int binary_search(int k){
int l = 1 , r = cnt;
while(l < r){
int mid = (l+r)/2;
if(sum[mid] >= k){
r = mid;
}else{
l = mid+1;
}
}
return l;
}

void computing(){
build(1 , cnt , 1);
int i , j , ans;
while(q--){
//cin >> i >> j;
scanf("%d%d" , &i ,&j);
int L = binary_search(i);
int R = binary_search(j);
if(L == R) printf("%d\n" , j-i+1);
else{
ans = max(sum[L]-i+1 , j-sum[R-1]);
if(L+1 <= R-1) ans = max(ans , querry(L+1 , R-1 , 1));
printf("%d\n" , ans);
}
}
}

int main(){
while(cin >> n && n){
initial();
readcase();
computing();
}
return 0;
}

[/code]