2019 Multi-University Training Contest 4

主席树 + 二分答案

对于这种区间内的值域问题一般用主席树进行求解。

因为数据范围只有1e6，所以不用离散化，直接建树即可。

题目要求找到区间内离p第k近的数，可以想到，这个问题具有单调性（某个区间长度有大于k个值，那么比这个区间长度更长的比如也有大于k个值）

于是我们可以二分答案这个距离，枚举离p距离为mid的范围(max(1, p - mid), min(p + mid, 1e6)

这样我们每次主席树上询问这个值域范围内的个数是否大于等于k判断枚举的mid是否可行，最后mid停下来的最小值就是答案。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define __fastIn ios::sync_with_stdio(false), cin.tie(0)
#define pb push_back
using namespace std;
typedef long long LL;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        ret = (ret << 3) + (ret << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -ret : ret;
}
inline int lcm(int a, int b){ return a / __gcd(a, b) * b; }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 200005;
const int M = 1000005;
int _, n, m, tot, a[N], lc[M*20], rc[M*20], tree[M*20], root[N];

int buildTree(int l, int r){
    int cur = ++ tot;
    if(l == r) return cur;
    int mid = (l + r) >> 1;
    buildTree(l, mid);
    buildTree(mid + 1, r);
    return cur;
}

int insert(int rt, int l, int r, int val){
    int cur = ++ tot;
    tree[cur] = tree[rt] + 1, lc[cur] = lc[rt], rc[cur] = rc[rt];
    if(l == r) return cur;
    int mid = (l + r) >> 1;
    if(val <= mid) lc[cur] = insert(lc[rt], l, mid, val);
    else rc[cur] = insert(rc[rt], mid + 1, r, val);
    return cur;
}

int query(int a, int b, int l, int r, int ql, int qr){
    if(l == ql && r == qr){
        return tree[b] - tree[a];
    }
    int mid = (l + r) >> 1;
    if(qr <= mid) return query(lc[a], lc[b], l, mid, ql, qr);
    else if(ql > mid) return query(rc[a], rc[b], mid + 1, r, ql, qr);
    return query(lc[a], lc[b], l, mid, ql, mid) + query(rc[a], rc[b], mid + 1, r, mid + 1, qr);
}

bool calc(int l, int r, int p, int k, int mid){
    int ret = query(root[l - 1], root[r], 1, M, max(1, p - mid), min(p + mid, M));
    return ret >= k;
}

int main(){

    //freopen("data.txt", "r", stdin);
    for(_ = read(); _; _ --){
        tot = 0;
        n = read(), m = read();
        for(int i = 1; i <= n; i ++) a[i] = read();
        root[0] = buildTree(1, M);
        for(int i = 1; i <= n; i ++){
            root[i] = insert(root[i - 1], 1, M, a[i]);
        }
        int ans = 0;
        while(m --){
            int L = read(), R = read(), p = read(), k = read();
            L ^= ans, R ^= ans, p ^= ans, k ^= ans;
            int l = 0, r = 1e6;
            while(l < r){
                int mid = (l + r) >> 1;
                if(calc(L, R, p, k, mid)) r = mid;
                else l = mid + 1;
            }
            printf("%d
", l);
            ans = l;
        }
    }
    return 0;
}