POJ 3368 Frequent values (UVA 11235)(RMQ)

畫船聽雨發表於2014-02-27
MQ

感覺不太好做吧，演算法訓練指南P198 UVA11235和這道題的解釋一樣。大家可以參考一下。

Frequent values
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 12436   Accepted: 4554

Description

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

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

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
3
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
#define M 10001000
#define LL __int64
//#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
const int maxn = 101000;

using namespace std;

int sum[maxn];
int f[maxn];
int l[maxn];
int r[maxn];
int dp[maxn][20];

int main()
{
    int n, m;
    while(~scanf("%d",&n))
    {
        if(!n)
            break;
        scanf("%d",&m);
        int t = 1;
        int L = 1, R = 1;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&f[i]);
            if(f[i] == f[i-1] && i != 1)
            {
                R++;
                t++;
            }
            if((f[i] != f[i-1] && i != 1) || i == n)
            {
                for(int j = L; j <= R; j++)
                {
                    sum[j] = t;
                    l[j] = L;
                    r[j] = R;
                }
                t = 1;
                L = R = i;
            }
        }
        for(int j = L; j <= n; j++)
        {
            sum[j] = t;
            l[j] = L;
            r[j] = n;
        }
        memset(dp, 0 , sizeof(dp));
        for(int i = 1; i <= n; i++)
            dp[i][0] = sum[i];
        for(int j = 1; (1 << j) <= n; j++)
            for(int i = 1; i+(1<<j)-1 <= n; i++)
                dp[i][j] = max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
        int x, y;
        while(m--)
        {
            scanf("%d %d",&x, &y);
            if(x > y)
                swap(x, y);
            L = r[x];
            R = l[y];
            if(L >= y)
            {
                printf("%d\n",y-x+1);
                continue;
            }
            if(R <= x)
            {
                printf("%d\n",y-x+1);
                continue;
            }
            int ans = max(L-x+1, y-R+1);
            if(L == R-1)
            {
                printf("%d\n",ans);
                continue;
            }
            L++;
            R--;
            int k=(int)(log(double(R-L+1))/log((double)2));
            ans = max(ans, max(dp[L][k], dp[R-(1<<k)+1][k]));
            printf("%d\n",ans);
        }
    }
    return 0;
}


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