P2048 [NOI2010]超级钢琴 (RMQ,堆)

大意: 给定n元素序列a, 定义一个区间的权值为区间内所有元素和, 求前k大的长度在[L,R]范围内的区间的权值和.

固定右端点, 转为查询左端点最小的前缀和, 可以用RMQ O(1)查询.

要求的是前$k$大, 可以用堆维护可供选择的区间, 每次取出最大的即可

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//head




const int N = 1e6+10;
int n, k, L, R;
int Log[N], a[N], f[N][20];
int RMQ(int l, int r) {
	int k=Log[r-l+1];
	int x=f[l][k],y=f[r-(1<<k)+1][k];
	return a[x]<a[y]?x:y;
}

struct _ { 
	int l,r,opt,pos,v;
	_ () {}
	_ (int l,int r,int pos) : l(l),r(r),pos(pos) {
		opt = RMQ(l,r);
		v = a[pos]-a[opt];
	}
	bool operator < (const _ & rhs) const {
		return v < rhs.v;
	}
};
priority_queue<_> q;

int main() {
	scanf("%d%d%d%d", &n, &k, &L, &R);
	REP(i,1,n) scanf("%d",a+i),a[i]+=a[i-1];
	Log[0]=-1;
	REP(i,1,n) f[i][0]=i,Log[i]=Log[i>>1]+1;
	for (int j=1; (1<<j)<=n; ++j) {
		for (int i=0; i+(1<<j)-1<=n; ++i) {
			int x=f[i][j-1],y=f[i+(1<<(j-1))][j-1];
			f[i][j]=a[x]<a[y]?x:y;
		}
	}
	REP(i,L,n) q.push(_(max(0,i-R),i-L,i));
	ll ans = 0;
	REP(i,1,k) {
		_ t = q.top();q.pop();
		ans += t.v;
		if (t.opt!=t.l) q.push(_(t.l,t.opt-1,t.pos));
		if (t.opt!=t.r) q.push(_(t.opt+1,t.r,t.pos));
	}
	printf("%lld
", ans);
}