【POJ3419】Difference Is Beautiful

最新推荐文章于 2021-03-27 10:37:42 发布

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最新推荐文章于 2021-03-27 10:37:42 发布

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分类专栏： RMQ 经典思维 POJ 文章标签： RMQ 思维

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本文介绍了一个算法问题，即如何找出一组公司营业额数据中，最长的没有重复元素的连续子序列长度。通过使用单调性、二分查找及区间最值查询等技术手段，提供了一种高效求解方法。

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Difference Is Beautiful

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 2247		Accepted: 703

Description

Mr. Flower's business is growing much faster than originally planned. He has now become the CEO of a world-famous beef corporation. However, the boss never lives a casual life because he should take charge of the subsidiary scattered all over the world. Every year, Mr. Flower needs to analyze the performance reports of these subsidiary companies.

Mr. Flower has N companies, and he numbered them with 0 to N – 1. All of the companies will give Mr. Flower a report about the development each year. Among all of the tedious data, only one thing draws Mr. Flower's attention – the turnover. Turnover of a company can be represented as an integer P_i: positive one represents the amount of profit-making while negative for loss-making.

In fact, Mr. Flower will not be angry with the companies running under deficit. He thinks these companies have a large room for future development. What dissatisfy him are those companies who created the same turnover. Because in his eyes, keeping more than one companies of the same turnover is not necessary.

Now we know the annual turnover of all companies (an integer sequence P_i, the ith represents the turnover of the ith company this year.). We say a number sequence is perfect if all of its numbers are different from each other. Mr. Flower wants to know the length of the longest consecutive perfect sequence in a certain interval [L, R] of the turnover sequence, can you help him?

Input

The first line of the input contains two integers N and M. N is the number of companies. M is the number of queries. (1 ≤ N, M ≤ 200000). The second line contains N integer numbers not exceeding 10⁶ by their absolute values. The ith of them represents the turnover of the ith company this year. The following M lines contain query descriptions, each description consists of two numbers: L, R (0 ≤ L ≤ R ≤ N – 1) and represents the interval that Mr. Flower concerned.

Output

The output contains M lines. For each query, output the length of the longest consecutive perfect sequence between [L, R] 　

Sample Input

9 2
2 5 4 1 2 3 6 2 4
0 8
2 6

Sample Output

6
5

Hint

The longest perfect sequence of the first query in the sample input is '5 4 1 2 3 6', so the answer for this query is 6.

Source

POJ Monthly--2007.10.06, SHOIT@ZSU

单调性+二分+RMQ；

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#define M 1000000
using namespace std;
const int N = 200000+10;
int la[M*2+100],dp[N],a[N],mm[N][25],mi[N][25];
int n,m;
void RMQ() {
	for(int i=0; i<n; i++) {
		mm[i][0]=mi[i][0]=la[i];
	}
	for(int j=1; j<20; j++) {
		for(int i=0; i<n; i++) {
			if(i+(1<<j)-1<n) {
				mm[i][j]=max(mm[i][j-1],mm[i+(1<<(j-1))][j-1]);
				mi[i][j]=min(mi[i][j-1],mi[i+(1<<(j-1))][j-1]);
			}
		}
	}
}
int max_rmq(int x,int y) {
	int cnt=(int)(log((double)(y-x+1))/log(2.0));
	return max(mm[x][cnt],mm[y-(1<<cnt)+1][cnt]);
}
int main() {
	while(scanf("%d%d",&n,&m)!=EOF) {
		for(int i=0; i<n; i++) {
			scanf("%d",&a[i]);
		}
		memset(la,-1,sizeof(la));
		la[a[0]+M]=0;
		dp[0]=0;
		for(int i=1; i<n; i++) {
			if(dp[i-1]>la[a[i]+M])
				dp[i]=dp[i-1];
			else
				dp[i]=la[a[i]+M]+1;
			la[a[i]+M]=i;
		}
		for(int i=0; i<n; i++) {
			la[i]=i-dp[i]+1;
		}
		RMQ();
		int x,y;
		while(m--) {
			scanf("%d%d",&x,&y);
			int temp=lower_bound(dp+x,dp+y+1,x)-dp;
			temp--;
			int ans=temp-x+1;
			if(temp+1<=y) {
				ans=max(ans,max_rmq(temp+1,y));
			}
			printf("%d\n",ans);
		}
	}
	return 0;
}

题目地址：http://poj.org/problem?id=3419