HDU 6534 Chika and Friendly Pairs (树状数组 离散化 莫队)

树状数组 离散化 莫队

题目链接

求[L,R]内有多少对(i, j)满足|a[i] - a[j]| <= K，我们转换为求[a[i] - k - 1, a[i] + k]的个数

离散化处理a[i], a[i] - k, a[i] + k的下标(二分处理)

然后莫队查询，用树状数组维护这些数，实现区间查询，区间修改

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#define rep(i, s, e) for(int i = s; i < e; ++i)
#define P pair<int, int>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
static const int N = 35;
static const int MAX_N = 27005;
static const ll Mod = 1e9 + 7;
int a[MAX_N], b[MAX_N * 3], c[MAX_N * 3], suml[MAX_N], sumr[MAX_N], rev[MAX_N];
int maxv;
struct Query{
    int l, r, pos, id;
    bool operator < (const Query &rhs)const{
        return (pos ^ rhs.pos) ? l < rhs.l : (pos & 1) ? r < rhs.r : r > rhs.r;
    }
}q[MAX_N];
int lowbit(int x){return x & (-x);}
void update(int p, int v){
    for(int i = p; i <= maxv; i += lowbit(i)) c[i] += v;
}
int query(int p){
    int rt = 0;
    for(int i = p; i; i -= lowbit(i)) rt += c[i];
    return rt;
}
void solve(){
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    maxv = n * 3;
    int cnt = 0;
    for(int i = 1; i <= n; ++i){
        scanf("%d", &a[i]);
        b[++cnt] = a[i] - k;
        b[++cnt] = a[i];
        b[++cnt] = a[i] + k;
    }
    int block = (int)sqrt(n);
    for(int i = 1; i <= m; ++i){
        scanf("%d%d", &q[i].l, &q[i].r);
        q[i].id = i;
        q[i].pos = (q[i].l - 1) / block + 1;
    }
    sort(b + 1, b + cnt + 1);
    int cur = unique(b + 1, b + cnt + 1) - b - 1;
    for(int i = 1; i <= n; ++i){
        suml[i] = lower_bound(b + 1, b + cur + 1, a[i] - k) - b;
        sumr[i] = lower_bound(b + 1, b + cur + 1, a[i] + k) - b;
        a[i] = lower_bound(b + 1, b + cur + 1, a[i]) - b;
    }
    sort(q + 1, q + m + 1);
    int l = 1, r = 0;
    int ans = 0;
    for(int i = 1; i <= m; ++i){
        while(l < q[i].l){
            update(a[l], -1);
            ans -= query(sumr[l]) - query(suml[l] - 1);
            ++l;
        }
        while(l > q[i].l){
            --l;
            ans += query(sumr[l]) - query(suml[l] - 1);
            update(a[l], 1);
        }
        while(r > q[i].r){
            update(a[r], -1);
            ans -= query(sumr[r]) - query(suml[r] - 1);
            --r;
        }
        while(r < q[i].r){
            ++r;
            ans += query(sumr[r]) - query(suml[r] - 1);
            update(a[r], 1);
        }
        rev[q[i].id] = ans;
    }
    for(int i = 1; i <= m; ++i) printf("%d\n", rev[i]);
}
int main() {
    solve();
    return 0;
}