Educational Codeforces Round 7 CF622C Not Equal on a Segment

最新推荐文章于 2019-01-24 08:54:39 发布

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本文介绍了一种高效算法，用于解决给定数组及多个查询的问题，每个查询需找出指定区间内与给定值不等的元素下标。通过预处理数组存储相等与不等值下标，实现快速查询。

Not Equal on a Segment
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given array a with n integers and m queries. The i-th query is given with three integers li, ri, xi.

For the i-th query find any position pi (li ≤ pi ≤ ri) so that api ≠ xi.

Input
The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the number of elements in a and the number of queries.

The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.

Each of the next m lines contains three integers li, ri, xi (1 ≤ li ≤ ri ≤ n, 1 ≤ xi ≤ 106) — the parameters of the i-th query.

Output
Print m lines. On the i-th line print integer pi — the position of any number not equal to xi in segment [li, ri] or the value - 1 if there is no such number.

Sample test(s)
input
6 4
1 2 1 1 3 5
1 4 1
2 6 2
3 4 1
3 4 2
output
2
6
-1
4

题意：
顾题思义，即给出一组数，并给出m个查询，每个查询分别由l,r,x三个数组成，表示在[l，r]范围内，求出与x值不相等的数的下标。
思路：
定义一个新数组sss存放原数组str中相等值与不等值的下标情况，sss[n] = n；当后一个数与当前数相等时，则下标与之一致，反之，当前数存放自身的下标。查询时效率就大大提高了。当str[l]的值不为x时，则输出该下标，否则，将sss[l]的值 + 1，如果此时的值小于r，则输出sss[l] + 1，否则输出-1，done。

#include<bits/stdc++.h>
using namespace std;
int main ()
{
    int n,m;
    while(cin>>n>>m)
    {
        int str[4400000];
        for(int i = 1;i <= n;i++)
            scanf("%d",&str[i]);
        int sss[4400000];
        sss[n] = n;
        for(int i = n - 1;i >= 0;i--)
        {
            if(str[i] == str[i + 1])
                sss[i] = sss[i + 1];
            else
                sss[i] = i;
        }
        while(m--)
        {
            int l,r,x;  scanf("%d %d %d",&l,&r,&x);
            if(str[l] != x)
                printf("%d\n",l);
            else
            {
                if(sss[l] + 1 <= r)
                    printf("%d\n",sss[l] + 1);
                else
                    printf("-1\n");
            }
        }
    }
    return 0;
}