HDU 5289 Assignment rmq

Assignment

题目连接：

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

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],…,an,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

题意

给你n个数，然后连续的，且这个区间内最大值减去最小值的差小于k的话，就可以算作一队。

问你一共有多少种分队的方法。

题解：

暴力枚举左端点位置，右端点单调递增，用一个rmq维护一下就好了。

Debug

mm[0]=-1;未赋初值

rmq(int x,int y);未般判断x>y的情况。

j+(1<<(i-1))和　j+1<<(i-1) 不同

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 100050
ll n,m,val[N],mx[N][20],mi[N][20],mm[N];
template<typename T>void read(T&x)
{
    ll k=0; char c=getchar();
    x=0;
    while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
    if (c==EOF)exit(0);
    while(isdigit(c))x=x*10+c-'0',c=getchar();
    x=k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c=getchar())&&c!=EOF);}
void init_ST(ll n)
{
    mm[0]=-1;//
    for(ll i=1;i<=n;i++)
    {
        mx[i][0]=mi[i][0]=val[i];
        mm[i]=(i&(i-1))==0?mm[i-1]+1:mm[i-1];
    }
    for(ll i=1;i<=20;i++)
        for(ll j=1;j+(1<<i)-1<=n;j++)
        {
            mx[j][i]=max(mx[j][i-1],mx[j+(1<<(i-1))][i-1]);
            mi[j][i]=min(mi[j][i-1],mi[j+(1<<(i-1))][i-1]);
        }
}
ll rmq(ll x,ll y)
{
    if(x>y)swap(x,y);//
    ll k=mm[y-x+1];
    ll ans=max(mx[x][k],mx[y-(1<<k)+1][k]);
    ans-=min(mi[x][k],mi[y-(1<<k)+1][k]);
    return ans;
}
void work()
{
    ll k;
    read(n); read(k);
    for(ll i=1;i<=n;i++)read(val[i]);
    init_ST(n);
    ll r=0,ans=0;;
    for(ll i=1;i<=n;i++)
    {
        while (r+1<=n&&rmq(i,r+1)<k)r++;
        ans+=r-i+1;
    }
    printf("%lld\n",ans);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("aa.in","r",stdin);
#endif
    ll q;
    read(q);
    while(q--)
    {
        work();
    }
}