题目链接
题解
只会做\(70\)分的\(O(nlog^2n)\)
如果本来就在区间内的人是不用动的，区间右边的人往区间最右的那些空位跑，区间左边的人往区间最左的那些空位跑
找到这些空位就用二分 + 主席树
理应可以在主席树上的区间二分而做到\(O(nlogn)\)，但是写不出来，先留着坑
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define REP(i,n) for (register int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,long long int>(a,b)
#define cp pair<int,long long int>
#define LL long long int
using namespace std;
const int maxn = 500005,maxm = 11000005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int N,n,m,rt[maxn];
int ls[maxm],rs[maxm],num[maxm],cnt;
LL sum[maxm];
void modify(int& u,int pre,int l,int r,int pos){
u = ++cnt;
sum[u] = sum[pre] + pos; num[u] = num[pre] + 1;
ls[u] = ls[pre]; rs[u] = rs[pre];
if (l == r) return;
int mid = l + r >> 1;
if (mid >= pos) modify(ls[u],ls[pre],l,mid,pos);
else modify(rs[u],rs[pre],mid + 1,r,pos);
}
int q_num(int u,int v,int l,int r,int L,int R){
if (l >= L && r <= R) return num[u] - num[v];
int mid = l + r >> 1;
if (mid >= R) return q_num(ls[u],ls[v],l,mid,L,R);
if (mid < L) return q_num(rs[u],rs[v],mid + 1,r,L,R);
return q_num(ls[u],ls[v],l,mid,L,R) + q_num(rs[u],rs[v],mid + 1,r,L,R);
}
LL q_sum(int u,int v,int l,int r,int L,int R){
if (l >= L && r <= R) return sum[u] - sum[v];
int mid = l + r >> 1;
if (mid >= R) return q_sum(ls[u],ls[v],l,mid,L,R);
if (mid < L) return q_sum(rs[u],rs[v],mid + 1,r,L,R);
return q_sum(ls[u],ls[v],l,mid,L,R) + q_sum(rs[u],rs[v],mid + 1,r,L,R);
}
inline LL S(int l,int r){
return 1ll * (l + r) * (r - l + 1) / 2;
}
inline LL q_pre(int u,int v,int L,int R,int k){
int ll = L,rr = R,mid; LL a;
while (ll < rr){
mid = ll + rr >> 1;
a = q_num(u,v,1,N,L,mid);
if ((mid - L + 1) - a >= k) rr = mid;
else ll = mid + 1;
}
a = q_sum(u,v,1,N,L,ll);
return S(L,ll) - a;
}
inline LL q_post(int u,int v,int L,int R,int k){
int ll = L,rr = R,mid,a;
while (ll < rr){
mid = ll + rr + 1 >> 1;
a = q_num(u,v,1,N,mid,R);
if ((R - mid + 1) - a >= k) ll = mid;
else rr = mid - 1;
}
a = q_sum(u,v,1,N,mid,R);
return S(ll,R) - a;
}
void work(){
int l,r,L,R,a,s; LL ans,b;
while (m--){
l = read(); r = read(); L = read(); R = L + r - l; ans = 0;
if (L > 1){
a = q_num(rt[r],rt[l - 1],1,N,1,L - 1);
if (a){
s = q_sum(rt[r],rt[l - 1],1,N,1,L - 1);
b = q_pre(rt[r],rt[l - 1],L,R,a);
ans += b - s;
}
}
a = q_num(rt[r],rt[l - 1],1,N,R + 1,N);
if (a){
s = q_sum(rt[r],rt[l - 1],1,N,R + 1,N);
b = q_post(rt[r],rt[l - 1],L,R,a);
ans += s - b;
}
printf("%lld\n",ans);
}
}
int main(){
n = read(); m = read(); N = 1000000 + n + 1; int x;
REP(i,n){
x = read(),modify(rt[i],rt[i - 1],1,N,x);
}
work();
return 0;
}