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 '\n'

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\n", ans);

}