hdu5172---GTY's gay friends

Problem Description

GTY has n gay friends. To manage them conveniently, every morning he ordered all his gay friends to stand in a line. Every gay friend has a characteristic value ai , to express how manly or how girlish he is. You, as GTY’s assistant, have to answer GTY’s queries. In each of GTY’s queries, GTY will give you a range [l,r] . Because of GTY’s strange hobbies, he wants there is a permutation [1..r?l+1] in [l,r]. You need to let him know if there is such a permutation or not.

Input

Multi test cases (about 3) . The first line contains two integers n and m ( 1≤n,m≤1000000 ), indicating the number of GTY’s gay friends and the number of GTY’s queries. the second line contains n numbers seperated by spaces. The ith number ai ( 1≤ai≤n ) indicates GTY’s ith gay friend’s characteristic value. The next m lines describe GTY’s queries. In each line there are two numbers l and r seperated by spaces ( 1≤l≤r≤n ), indicating the query range.

Output

For each query, if there is a permutation [1..r?l+1] in [l,r], print ‘YES’, else print ‘NO’.

Sample Input

8 5 2 1 3 4 5 2 3 1 1 3 1 1 2 2 4 8 1 5 3 2 1 1 1 1 1 1 2

Sample Output

YES NO YES YES YES YES NO

Source

BestCoder Round #29

Recommend

hujie | We have carefully selected several similar problems for you: 5173 5169 5168 5165 5164

如果一个区间里是一个[1, … , r - l + 1]的排列，那么首先，区间和是

x * (x + 1) / 2， –>前缀和处理

其次每一个数都不同，预处理每一个数上一次出现的位置pre，然后求出区间里pre的最大值，如果最大的pre小于左端点且区间和是 x * (x + 1) / 2，那么输出YES，否则NO，这里可以用线段树解决

注意判断时，最好先判断区间和，如果这里已经不满足了，就别去查询线段树了，否则容易造成TLE

/*************************************************************************
    > File Name: hdu5172.cpp
    > Author: ALex
    > Mail: [email protected]
    > Created Time: 2015年02月14日 星期六 12时49分13秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int N = 1000010;
int pre[N];
LL sum[N];
int _hash[N];

struct node
{
    int l, r;
    int val;
}tree[N << 2];

void build (int p, int l, int r)
{
    tree[p].l = l;
    tree[p].r = r;
    if (l == r)
    {
        tree[p].val = pre[l];
        return;
    }
    int mid = (l + r) >> 1;
    build (p << 1, l, mid);
    build (p << 1 | 1, mid + 1, r);
    tree[p].val = max (tree[p << 1].val, tree[p << 1 | 1].val);
}

int query (int p, int l, int r)
{
    if (tree[p].l == l && tree[p].r == r)
    {
        return tree[p].val;
    }
    int mid = (tree[p].l + tree[p].r) >> 1;
    if (r <= mid)
    {
        return query (p << 1, l, r);
    }
    else if (l > mid)
    {
        return query (p << 1 | 1, l, r);
    }
    else
    {
        return max (query (p << 1, l, mid), query (p << 1 | 1, mid + 1, r));
    }
}

int main ()
{
    int n, m;
    while (~scanf("%d%d", &n, &m))
    {
        memset (_hash, -1, sizeof(_hash));
        memset (sum, 0, sizeof(sum));
        int x, l, r;
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", &x);
            pre[i] = _hash[x];
            _hash[x] = i;
            sum[i] = sum[i - 1] + (LL)x;
        }
        build (1, 1, n);
        while (m--)
        {
            scanf("%d%d", &l, &r);
            LL len = (LL)(r - l + 1);
            len = (len + 1) * len / 2;
            if (len != sum[r] - sum[l - 1])
            {
                printf("NO\n");
                continue;
            }
            int res = query (1, l, r);
            if (res < l)
            {
                printf("YES\n");
            }
            else
            {
                printf("NO\n");
            }
        }
    }
    return 0;
}

hdu5172---GTY's gay friends