BZOJ-3289 Mato的文件管理

用分块莫队离线询问的同时用树状数组快速计算逆序对个数。

#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <fstream>
#include <iostream>
#include <queue>

#define rep(i, l, r) for(int i = l; i <= r; i++)
#define down(i, l, r) for(int i = l; i >= r; i--)
#define N 56789
#define ll long long

using namespace std;
int read()
{
	int x = 0, f = 1; char ch = getchar();
	while (ch<'0' || ch>'9') { if (ch=='-') f = -1; ch = getchar(); }
	while (ch>='0' && ch<='9') { x = x*10 + ch-'0'; ch = getchar(); }
	return x*f;
}
struct node{int l, r, id;} q[N];
int n, m, k[N], a[N], pos[N], ans[N], s[N], now;

bool cmp(node a, node b)
{
	if (pos[a.l] == pos[b.l]) return a.r < b.r; else return a.l < b.l;
}

int Q(int x)
{
	int a = 0; while (x>0) a+=s[x], x-=x&(-x); return a;
}

void Add(int x, int c) { while (x<=n) s[x]+=c, x+=x&(-x); }

int main()
{
	n=read(); 
	rep(i, 1, n) k[i]=a[i]=read();
	sort(a+1, a+1+n); 
	rep(i, 1, n) k[i] = lower_bound(a+1, a+n+1, k[i])-a;
	int block = int(sqrt(n)); 
	rep(i, 1, n) pos[i] = (i-1)/block+1;
	m=read(); 
	rep(i, 1, m) q[i].l=read(), q[i].r=read(), q[i].id = i;
	sort(q+1, q+1+m, cmp);
	int l = 1, r = 0;
	rep(i, 1, m)
	{
		while (l < q[i].l) 
			now -= Q(k[l]-1), Add(k[l], -1), l++;
		while (l > q[i].l) 
			l--, now += Q(k[l]-1), Add(k[l], 1);
		while (r < q[i].r) 
			r++, now += r-l-Q(k[r]), Add(k[r], 1);
		while (r > q[i].r) 
			now -= r-l+1-Q(k[r]), Add(k[r], -1), r--;
		ans[q[i].id] = now;
	}
	rep(i, 1, m) printf("%d\n", ans[i]);
	return 0;
}