HDU 5289 Assignment [优先队列 贪心]

HDU 5289 - Assignment

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

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:a1,a2,…,an(0 <= ai <= 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

这题就是给定一个序列，求出满足区间内最大值最小值之差小于k的子序列的数量。

我是用优先队列每次更新区间左端点，一遍for循环写的。队里的dalao是用RMQ先预处理完然后二分做的。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) (x&(-x))
#define eps 1e-8
using namespace std;
typedef long long ll;

const int maxn = 1e5+7;
int arr[maxn],n,k;

struct node1{
int pos,val;
node1(int _pos,int _val):pos(_pos),val(_val){}
bool operator < (const node1 &c) const{
return (val < c.val);
}
};
struct node2{
int pos,val;
node2(int _pos,int _val):pos(_pos),val(_val){}
bool operator < (const node2 &c) const{
return (val > c.val);
}
};
priority_queue <node1> high;
priority_queue <node2> low;

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
while(!high.empty()) high.pop();
while(!low.empty()) low.pop();
scanf("%d%d",&n,&k);
for(int i=0;i<n;i++)
scanf("%d",&arr[i]);
ll ans = 1;
int l = 0, MIN = INF, MAX = -INF;//, pmi = 0, pma = 0;
for(int i=0;i<n;i++)
{
MIN = min(MIN,arr[i]);
MAX = max(MAX,arr[i]);
high.push(node1(i,arr[i]));
low.push(node2(i,arr[i]));
if(MAX - MIN >= k)
{
if(arr[i] == MAX)
{
while(!low.empty() && arr[i] - MIN >= k)
{
l = max(l,low.top().pos + 1);
low.pop();
MIN = low.top().val;
}
}
else
{
while(!high.empty() && MAX - arr[i] >= k)
{
l = max(high.top().pos + 1,l);
high.pop();
MAX = high.top().val;
}
}
}
ans += i-l+1;
}
printf("%lld\n",ans-1);
}
}