hdu3530Subsequence【单调队列优化dp】2010多校联合

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Problem Description

There is a sequence of integers. Your task is to find the longest subsequence that satisfies the following condition: the difference between the maximum element and the minimum element of the subsequence is no smaller than m and no larger than k.

Input

There are multiple test cases.
For each test case, the first line has three integers, n, m and k. n is the length of the sequence and is in the range [1, 100000]. m and k are in the range [0, 1000000]. The second line has n integers, which are all in the range [0, 1000000].
Proceed to the end of file.

Output

For each test case, print the length of the subsequence on a single line.

Sample Input

5 0 01 1 1 1 15 0 31 2 3 4 5

Sample Output

Source

2010 ACM-ICPC Multi-University Training Contest（10）——Host by HEU

低估了多校的难度，高估了自己的实力==从下午4点多弄到现在，勉勉强强写出了了

说题意，给定一个数列，求最大值与最小值大于等于m小于等于k的连续最长长度，想到用单调队列优化最值问题，类似于

FZU - 1894选拔志愿者【单调队列】

的写法，只不过要维护两个数组分别记录递增的最小值和递减的最大值，又如同

hdu3401trade【单调队列优化dp】

我们需要根据已知条件删除队首元素，我们要求最值差不大于k,不小于m，但是队首元素相减只能剪出来最值差的最大值，那索性就用最大差值来控制队首元素呗，那最小差值怎么办？取最优解的时候不用不就完了嘛。还有就是区间长度不是由寻找到的最右边-最左边求出的，而是当前位置减去最左边求出的，最左边一定存在与两个单调队列的队首之一！要不我怎么说他像选拔志愿者那个题呢QAQ队首元素一边用k值控制，控制到满足题意就是最左区间啦

/*****************hdu35302016.2.23156MS2900K1107 BC++******************/#include <iostream>#include<cstdio>#include<cstring>using namespace std;int n,m,k,num[100005],qmin[100005],qmax[100005],dp;int lmin,rmin,lmax,rmax;int max(int x,int y){    return x>y?x:y;}int main(){    //freopen("cin.txt","r",stdin);//单调队列只开一个！！！要不就乱套了    while(~scanf("%d%d%d",&n,&m,&k))    {        lmax=lmin=0,rmax=rmin=-1;        dp=0;        int l=0;        for(int i=1;i<=n;i++)        {            scanf("%d",&num[i]);            while(lmin<=rmin&&num[qmin[rmin]]>num[i]) rmin--;//方向居然能写反！            qmin[++rmin]=i;            while(lmax<=rmax&&num[qmax[rmax]]<num[i]) rmax--;            qmax[++rmax]=i;            while(num[qmax[lmax]]-num[qmin[lmin]]>k)                l=(qmax[lmax]<qmin[lmin])?qmax[lmax++]:qmin[lmin++];            if(num[qmax[lmax]]-num[qmin[lmin]]>=m&&num[qmax[lmax]]-num[qmin[lmin]]<=k)            dp=max(dp,i-l);//我在取值的时候只取了队首，所以不需要控制队尾元素！        }        printf("%d\n",dp);    }    return 0;}

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