hdu 5289 Assignment 二分+RMQ 2015 Multi-University Training Contest 1 02

题意：给出1～n的一个序列，求区间最大值-区间最小值<k的区间数。

RMQ处理出区间最大最小值，易知某段区间的最大值-其最小值会随着区间长度的减小而减小，因此可以通过枚举左端点，二分右端点来计算以左端点为起点的区间满足条件的贡献。

#include <bits/stdc++.h>#include <map>#include <set>#include <queue>#include <stack>#include <cmath>#include <time.h>#include <vector>#include <cstdio>#include <string>#include <iomanip>///cout << fixed << setprecision(13) << (double) x << endl;#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1 | 1#define ls rt << 1#define rs rt << 1 | 1#define pi acos(-1.0)#define eps 1e-8#define Mp(a, b) make_pair(a, b)#define asd puts("asdasdasdasdasdf");typedef long long ll;//typedef __int64 LL;const int inf = 0x3f3f3f3f;const int N = 101000;int dp1[N][20];int dp2[N][20];int LOG[N];int a[N];void init_LOG(){LOG[0] = -1;for( int i = 1; i <= N-10; ++i )LOG[i] = ( (i&(i-1) ) == 0 ) ? LOG[i-1] + 1 : LOG[i-1];}void init_RMQ( int n ){for( int i = 1; i <= n; ++i )dp1[i][0] = dp2[i][0] = a[i];for( int j = 1; j <= LOG[n]; ++j ) {for( int i = 1; i + ( 1 << j ) - 1 <= n; ++i ) {dp1[i][j] = max( dp1[i][j-1], dp1[i+(1<<(j-1))][j-1] );dp2[i][j] = min( dp2[i][j-1], dp2[i+(1<<(j-1))][j-1] );}}}int cal( int l, int r ){int k = LOG[r-l+1];int maxx = max( dp1[l][k], dp1[r-(1<<k)+1][k] );int minn = min( dp2[l][k], dp2[r-(1<<k)+1][k] );return maxx - minn;}int main(){init_LOG();int tot, k, n;for( scanf("%d", &tot); tot--; ) {scanf("%d%d", &n, &k);for( int i = 1; i <= n; ++i ) {scanf("%d", &a[i]);}init_RMQ(n);ll ans = 0;for( int i = 1; i <= n; ++i ) {ll l = i, r = n, mid;while( l <= r ) {mid = (l+r) >> 1;ll tmp = cal( i, mid );if( tmp < k )l = mid+1;elser = mid-1;}ll len = l - i;ans += len;}printf("%lld\n", ans);}return 0;}