hdu 5289 Assignment 二分+RMQ 2015 Multi-University Training Contest 1 02
2015-07-31 21:07
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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。
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 [20]; int dp2 [20]; int LOG ; int a ; 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 ; ++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; }
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