“玲珑杯”ACM比赛 Round #19 B -- Buildings （RMQ + 二分）

“玲珑杯”ACM比赛 Round #19

Start Time：2017-07-29 14:00:00 End Time：2017-07-29 16:30:00 Refresh Time：2017-07-29 16:42:55 Private

B -- Buildings

Time Limit：2s Memory Limit：128MByte

Submissions：590Solved：151

DESCRIPTION

There are $\mathrm{nn\; buildings\; lined\; up,\; and\; the\; height\; of\; the\; ii-th\; house\; is\; hihi.}$

An inteval $[l,r][l,r](l\le r)(l\le r)\; is\; harmonious\; if\; and\; only\; if\; max(hl,\dots ,hr)-min(hl,\dots ,hr)\le kmax(hl,\dots ,hr)-min(hl,\dots ,hr)\le k.$

Now you need to calculate the number of harmonious intevals.

INPUT

The first line contains two integers $\mathrm{n(1\le n\le 2\times 105),k(0\le k\le 109)n(1\le n\le 2\times 105),k(0\le k\le 109).\; The\; second\; line\; contains\; nn\; integers\; hi(1\le hi\le 109)hi(1\le hi\le 109).}$

OUTPUT

Print a line of one number which means the answer.

SAMPLE INPUT

3 1 1 2 3

SAMPLE OUTPUT

5

HINT

Harmonious intervals are: $[1,1],[2,2],[3,3],[1,2],[2,3][1,1],[2,2],[3,3],[1,2],[2,3].$

$【题意】给你一个序列，然后问你有多少个区间的最大值-最小值<=k。$

$【分析】我们可以用RMQ来logn查询区间最大、最小值，然后枚举每一位最为区间右端点，向左二分左端点，将区间长度加到ans即可。$

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 2e5+50;
const int mod = 1e9+7;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
int n,k;
int a[N];
int mn[N][20],mx[N][20],mm[N];
void init() {
    for(int j=1; j<=mm[n]; ++j) {
        for(int i=1; i+(1<<j)-1<=n; ++i) {
            mn[i][j]=min(mn[i][j-1],mn[i+(1<<(j-1))][j-1]);
            mx[i][j]=max(mx[i][j-1],mx[i+(1<<(j-1))][j-1]);
        }
    }
}
int getmx(int l,int r) {
    int k = mm[r-l+1];
    return max(mx[l][k],mx[r-(1<<k)+1][k]);
}
int getmn(int l,int r) {
    int k = mm[r-l+1];
    return min(mn[l][k],mn[r-(1<<k)+1][k]);
}
int main(){
    mm[0]=-1;
    for(int i=1; i<N; ++i)mm[i]=(i&(i-1))?mm[i-1]:mm[i-1]+1;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        mn[i][0]=mx[i][0]=a[i];
    }
    init();
    int l,r,res;
    ll ans=0;
    for(int i=1;i<=n;i++){
        l=1;r=i;res=i;
        while(l<=r){
            int mid=(l+r)/2;
            int maxn=getmx(mid,i);
            int minn=getmn(mid,i);
            if(maxn-minn>k)l=mid+1;
            else r=mid-1,res=mid;
        }
        ans+=i-res+1;
    }
    printf("%lld
",ans);
    return 0;
}