【LOJ】#2551. 「JSOI2018」列队

题解

老年选手一道裸的主席树都要看好久才看出来

首先熟练的把这个区间建成(n)个主席树
然后对于一个询问，我们相当于在主席树上二分一个mid，使得(mid - K + 1)正好和([l,r])区间中坐标在([1,mid])的人数一样就好

（居然代码只有2.2K，似乎比平均码长低啊，开心）

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('
')
#define space putchar(' ')
#define MAXN 500005
#define eps 1e-8
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 + c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MAXV = 2000000;
struct node {
    int lc,rc;
    int64 sum,cnt;
}tr[MAXN * 22];
int N,M;
int a[MAXN],rt[MAXN],Ncnt;
int64 f(int l,int r) {
    return 1LL * (l + r) * (r - l + 1) / 2;
}
void Insert(const int &x,int &y,int L,int R,int v) {
    y = ++Ncnt;
    tr[y] = tr[x];
    tr[y].sum += v;tr[y].cnt++;
    if(L == R) return;
    int mid = (L + R) >> 1;
    if(v <= mid) Insert(tr[x].lc,tr[y].lc,L,mid,v);
    else Insert(tr[x].rc,tr[y].rc,mid + 1,R,v);
}
pair<int64,int64> Query(int x,int y,int L,int R,int K,int len) {
    pair<int64,int64> res = mp(0,0);
    while(L <= R) {
	if(L == R) {
	    res.fi += tr[y].sum - tr[x].sum;
	    res.se += tr[y].cnt - tr[x].cnt;
	    break;
	}
	pair<int64,int64> t = res;
	int mid = (L + R) >> 1;
	t.fi += tr[tr[y].lc].sum - tr[tr[x].lc].sum;
	t.se += tr[tr[y].lc].cnt - tr[tr[x].lc].cnt;
	if(mid >= K && mid - K + 1 >= t.se) {
	    R = mid;
	    x = tr[x].lc;y = tr[y].lc;
	}
	else {
	    L = mid + 1;
	    res = t;
	    x = tr[x].rc;y = tr[y].rc;
	}
    }
    return res;
}
void Solve() {
    read(N);read(M);
    for(int i = 1 ; i <= N ; ++i) read(a[i]);
    for(int i = 1 ; i <= N ; ++i) {
	Insert(rt[i - 1],rt[i],1,MAXV,a[i]);
    }
    int l,r,K;
    for(int i = 1 ; i <= M ; ++i) {
	read(l);read(r);read(K);
	pair<int64,int64> t;
	t = Query(rt[l - 1],rt[r],1,MAXV,K,r - l + 1);
	int64 ans = f(K,K + t.se - 1) - f(K + t.se,K + r - l);
	ans -= t.fi;
	ans += tr[rt[r]].sum - tr[rt[l - 1]].sum - t.fi;
	out(ans);enter;
    }
    
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}