HDU 5289 Assignment（二分+RMQ-ST）

Posted 2020-08-31 Blackops

Assignment

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3813    Accepted Submission(s): 1771

Problem Description

Tom owns a company and he is the boss. There are n staffs which are numbered from 1 to n in this company, and every staff has a ability. Now, Tom is going to assign a special task to some staffs who were in the same group. In a group, the difference of the ability of any two staff is less than k, and their numbers are continuous. Tom want to know the number of groups like this.

 

 

Input

In the first line a number T indicates the number of test cases. Then for each case the first line contain 2 numbers n, k (1<=n<=100000, 0<k<=10^9),indicate the company has n persons, k means the maximum difference between abilities of staff in a group is less than k. The second line contains n integers:a[1],a[2],…,a[n](0<=a[i]<=10^9),indicate the i-th staff’s ability.

 

 

Output

For each test，output the number of groups.

 

 

Sample Input

2

4 2

3 1 2 4

10 5

0 3 4 5 2 1 6 7 8 9

 

 

Sample Output

5

28

Hint

First Sample, the satisfied groups include:[1,1]、[2,2]、[3,3]、[4,4] 、[2,3]

 

题目链接：HDU 5289

题意就是求有多少个连续子串，这些子串均符合相邻数之间差的绝对值均小于k，直接两个for计数估计会T，因此可以枚举子串的左端点$i$，二分右端点$R$，使得$[i,R]$长度最大，那么这样一来这个子串是肯定符合的，实际上把右端点往左缩一个得到的小一个单位的子串也肯定是符合的，这样可以一直缩到区间变成$[i,i]$，因此每一次枚举得到的区间$[i,R]$可以产生$R-i+1$个符合题意的子串。

另外由于最大答案可以达到$\frac{(1+10^5)*10^5} {2}$因此要用long long

代码：

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 1e5 + 7;
int arr[N], Max[N][20], Min[N][20];

void rmq_init(int l, int r)
{
    int i, j;
    for (i = l; i <= r; ++i)
        Max[i][0] = Min[i][0] = arr[i];
    for (j = 1; l + (1 << j) - 1 <= r; ++j)
    {
        for (i = l; i + (1 << j) - 1 <= r; ++i)
        {
            Max[i][j] = max(Max[i][j - 1], Max[i + (1 << (j - 1))][j - 1]);
            Min[i][j] = min(Min[i][j - 1], Min[i + (1 << (j - 1))][j - 1]);
        }
    }
}
pii ST(int l, int r)
{
    int k = log2(r - l + 1);
    int Ma = max(Max[l][k], Max[r - (1 << k) + 1][k]);
    int Mi = min(Min[l][k], Min[r - (1 << k) + 1][k]);
    return pii(Ma, Mi);
}
int main(void)
{
    int tcase;
    scanf("%d", &tcase);
    while (tcase--)
    {
        int n, k, i;
        scanf("%d%d", &n, &k);
        for (i = 1; i <= n; ++i)
            scanf("%d", &arr[i]);
        rmq_init(1, n);

        LL ans = 0;
        for (i = 1; i <= n; ++i)
        {
            int L = i, R = n;
            int idx = i;
            while (L <= R)
            {
                int mid = MID(L, R);
                pii temp = ST(i, mid);
                if (temp.first - temp.second < k)
                {
                    idx = mid;
                    L = mid + 1;
                }
                else
                    R = mid - 1;
            }
            ans += (LL)(idx - i + 1);
        }
        printf("%I64d\n", ans);
    }
    return 0;
}