Assignment
Time Limit: 4000/2000 MS (Java/Others) Memory
Limit: 65536/65536 K (Java/Others)
Total Submission(s): 627 Accepted Submission(s): 318
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.
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.
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
28HintFirst Sample, the satisfied groups include:[1,1]、[2,2]、[3,3]、[4,4] 、[2,3]
Author
FZUACM
Source
解题思路：
枚举左端点。二分查找右端点，用RMQ维护区间最大值和最小值
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <vector>
#include <stack>
#define LL long long
using namespace std;
const int MAXN = 100000 + 10;
int A[MAXN];
int dp1[MAXN][20], dp[MAXN][20];
int N, M;
void RMQ_init_Min()
{
for(int i=0;i<N;i++) dp1[i][0] = A[i];
for(int j=1;(1<<j) <= N;j++)
{
for(int i=0;i+(1<<j) - 1 < N;i++)
{
dp1[i][j] = min(dp1[i][j-1], dp1[i + (1<<(j-1))][j-1]);
}
}
}
int RMQ_Min(int L, int R)
{
int k = 0;
while(1<<(k+1) <= R - L + 1) k++;
return min(dp1[L][k], dp1[R-(1<<k)+1][k]);
}
void RMQ_init_Max()
{
for(int i=0;i<N;i++) dp[i][0] = A[i];
for(int j=1;(1<<j) <= N;j++)
{
for(int i=0;i+(1<<j) - 1 < N;i++)
{
dp[i][j] = max(dp[i][j-1], dp[i + (1<<(j-1))][j-1]);
}
}
}
int RMQ_Max(int L, int R)
{
int k = 0;
while(1<<(k+1) <= R - L + 1) k++;
return max(dp[L][k], dp[R-(1<<k)+1][k]);
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
scanf("%d%d", &N, &M);
for(int i=0;i<N;i++) scanf("%d", &A[i]);
RMQ_init_Min();
RMQ_init_Max();
long long ans = 0;
for(int i=0;i<N;i++)
{
int l = i, r = N-1;
while(l <= r)
{
int mid = (l + r) >> 1;
int low = RMQ_Min(i, mid);
int high = RMQ_Max(i ,mid);
//cout << l << ' ' << r << ' ' << high << ' ' <<low << endl;
if(high - low < M) l = mid + 1;
else r = mid - 1;
}
//cout << l << endl;
ans += (l - i);
}
printf("%I64d\n", ans);
}
return 0;
}