[2019杭电多校第四场][hdu6621]K-th Closest Distance（主席树）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6621

题意为求区间[l,r]内第k小|a[i]-p|的值。

可以二分答案，如果二分的值为x，则判断区间[l,r]内是否有k个数在[p-x,p+x]范围内。所以就用主席树搞一下。

 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<vector>
 #define lson l, mid, i<<1
 #define rson  mid + 1, r, i<<1|1
 using namespace std;
 typedef long long ll;
 const int maxn = 1e6 + ;
 int root[maxn], rs[maxn * ], ls[maxn * ], val[maxn * ];
 int cnt;
 void update(int k, int l, int r, int &i) {
     val[++cnt] = val[i] + , ls[cnt] = ls[i], rs[cnt] = rs[i];
     i = cnt;
     if (l == r)
         return;
     int mid = l + r >> ;
     if (k <= mid)
         update(k, l, mid, ls[i]);
     else
         update(k, mid + , r, rs[i]);
 }
 int query(int u, int v, int L, int R, int l, int r) {
     if (L <= l && r <= R)
         return val[u] - val[v];
     int mid = l + r >> ;
     int ans = ;
     if (L <= mid)
         ans += query(ls[u], ls[v], L, R, l, mid);
     if (R > mid)
         ans += query(rs[u], rs[v], L, R, mid + , r);
     return ans;
 }
 int check(int L, int R, int p, int x) {
     int l = max(p - x, ), r = min(p + x, (int)1e6);
     return query(root[R], root[L - ], l, r, , 1e6);
 }
 int main() {
     int t;
     scanf("%d", &t);
     while (t--) {
         int n, m, x, cnt = ;
         scanf("%d%d", &n, &m);
         for (int i = ; i <= n; i++) {
             scanf("%d", &x);
             root[i] = root[i - ];
             update(x, , 1e6, root[i]);
         }
         int ans = , L, R, p, k;
         for (int i = ; i <= m; i++) {
             scanf("%d%d%d%d", &L, &R, &p, &k);
             L ^= ans, R ^= ans, p ^= ans, k ^= ans;
             int l = , r = 1e6;
             while (l <= r) {
                 int mid = l + r >> ;
                 int w = check(L, R, p, mid);
                 if (w >= k) {
                     r = mid - ;
                     ans = mid;
                 }
                 else
                     l = mid + ;
             }
             printf("%d\n", ans);
         }
     }
 }