hdu 5289 Assignment 【ST算法】

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5289

题意：求满足最大值减最小值小于k的区间的数目。

枚举左端点，二分右端点，用st算法求区间最值

代码：

#include <stdio.h>
#include <ctime>
#include <math.h>
#include <limits.h>
#include <complex>
#include <string>
#include <functional>
#include <iterator>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <bitset>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <time.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>

using namespace std;

int n, k, p, l, r;

int a[100010];
int s[100010][20];
int maxsum[100010][20], minsum[100010][20];
__int64 ans = 0;

int rmq(int l, int r)
{
    int k = log2((double)(r - l + 1));
    int MAX = max(maxsum[l][k], maxsum[r - (1 << k) + 1][k]);
    int MIN = min(minsum[l][k], minsum[r - (1 << k) + 1][k]);
    return MAX - MIN;
}

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d %d", &n, &k);
        for (int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            maxsum[i][0] = minsum[i][0] = a[i];
        }

        for (int j = 1; (1 << j) <= n; j++)
            for (int i = 1; i + (1 << j) - 1 <= n; i++)
            {
                maxsum[i][j] = max(maxsum[i][j - 1], maxsum[i + (1 << (j - 1))][j - 1]);
                minsum[i][j] = min(minsum[i][j - 1], minsum[i + (1 << (j - 1))][j - 1]);
            }

        ans = 0;
        for (int i = 1; i <= n; i++)
        {
            l = i;
            r = n;

            while (l + 1 < r)
            {
                p = (l + r) >> 1;
                if (rmq(i, p) < k)
                    l = p;
                else
                    r = p;
            }
            if (rmq(i, r) < k)
                ans += (__int64)(r - i + 1);
            else
                ans += (__int64)(l - i + 1);
        }
        printf("%I64d\n", ans);
    }
    return 0;
}

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