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

最新推荐文章于 2016-06-13 21:48:36 发布

原创最新推荐文章于 2016-06-13 21:48:36 发布 · 295 阅读

0 ·

CC 4.0 BY-SA版权

HDU 专栏收录该内容

149 篇文章

订阅专栏

本文介绍了如何使用RMQ（线段树等）算法高效计算给定序列中区间最大值与最小值之差小于k的区间数量。通过枚举左端点并利用二分查找优化右端点搜索过程，实现复杂度优化。

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;
				else
					r = mid-1;
			}
			ll len = l - i;
			ans += len;
		}
		printf("%lld\n", ans);
	}
	return 0;
}