【二分+RMQ】玲珑oj 1149

1149 - Buildings

Time Limit：2s Memory Limit：128MByte

Submissions：820Solved：225

DESCRIPTION

There are $n$ buildings lined up, and the height of the $i$-th house is $h_i$.

An inteval $[l,r]$$(l\le r)$ is harmonious if and only if $max(h_l,\dots ,h_r)-min(h_l,\dots ,h_r)\le k$.

Now you need to calculate the number of harmonious intevals.

INPUT

The first line contains two integers$n(1\le n\le 2\times {10}^5),k(0\le k\le {10}^9)$.The second line contains $n$ integers $h_i(1\le h_i\le {10}^9)$.

OUTPUT

Print a line of one number which means the answer.

SAMPLE INPUT

3 11 2 3

SAMPLE OUTPUT

5

HINT

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

SOLUTION

“玲珑杯”ACM比赛 Round #19

题意：给定一个序列，问有多少个区间的最值差是小于等于k的；

思路：找到差值小于等于k的最大的区间。固定左端点，枚举由端点，因为区间越大，最大值就越大，最小值越小，即差值越大，具有单调性，所以二分右端点，找到满足差值小于等于k的最大区间,区间最值差用RMQ预处理出来；然后ans+=r-l+1，想一下为什么加上 r-l+1 ?

这道题目无限tle，原因是我把，maxx,minn数组定义成了 long long 型，但是看数据范围根本就没必要，所以特别是定义数组的时候不该用long long 就一定不要用，以防超时；

代码：

 

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;typedef long long ll;#define N 200010int a[N], lg[N], maxx[20][N], minn[20][N];int query(int l, int r){    int k = lg[r - l + 1];return max(maxx[k][l], maxx[k][r - (1 << k) + 1]) - min(minn[k][l], minn[k][r - (1 << k) + 1]);}void get_a(int n){    for(int i = 1; i <= n; i ++)        maxx[0][i] = minn[0][i] = a[i];    for(int i = 2; i <= n; i ++) if(i & (i - 1)) lg[i] = lg[i - 1];        else lg[i] = lg[i - 1] + 1;    for(int i = 1; (1 << i) <= n; i ++)    {        for(int j = 1; j + (1 << i) - 1 <= n; j ++)        {            maxx[i][j] = max(maxx[i - 1][j], maxx[i - 1][j + (1 << (i - 1))]);            minn[i][j] = min(minn[i - 1][j], minn[i - 1][j + (1 << (i - 1))]);        }    }}int main(){    int n, K;    scanf("%d%d", &n, &K);    for(int i = 1; i <= n; i ++)        scanf("%d", &a[i]);    get_a(n);    ll ans=0;    for(int i = 1; i <= n; i ++)    {        int l=i,r=n,mid;        while(l+1<r)        {            mid=(l+r)>>1;            if(query(i,mid)<=K)                l=mid;            else                r=mid;        }        if(query(i,r)<=K)            ans+=r-i+1;        else            ans+=l-i+1;    }    printf("%lld\n",ans);    return 0;}