HYSBZ3196-Tyvj 1730 二逼平衡树

Tyvj 1730 二逼平衡树

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 4293  Solved: 1652
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Description

您需要写一种数据结构（可参考题目标题），来维护一个有序数列，其中需要提供以下操作：
1.查询k在区间内的排名
2.查询区间内排名为k的值
3.修改某一位值上的数值
4.查询k在区间内的前驱(前驱定义为小于x，且最大的数)
5.查询k在区间内的后继(后继定义为大于x，且最小的数)

Input

第一行两个数 n,m 表示长度为n的有序序列和m个操作
第二行有n个数，表示有序序列
下面有m行，opt表示操作标号
若opt=1 则为操作1，之后有三个数l,r,k 表示查询k在区间[l,r]的排名
若opt=2 则为操作2，之后有三个数l,r,k 表示查询区间[l,r]内排名为k的数
若opt=3 则为操作3，之后有两个数pos,k 表示将pos位置的数修改为k
若opt=4 则为操作4，之后有三个数l,r,k 表示查询区间[l,r]内k的前驱
若opt=5 则为操作5，之后有三个数l,r,k 表示查询区间[l,r]内k的后继

Output

对于操作1,2,4,5各输出一行，表示查询结果

Sample Input

9 6
4 2 2 1 9 4 0 1 1
2 1 4 3
3 4 10
2 1 4 3
1 2 5 9
4 3 9 5
5 2 8 5

Sample Output

2
4
3
4
9

HINT

1.n和m的数据范围：n,m<=50000

2.序列中每个数的数据范围：[0,1e8]

3.虽然原题没有，但事实上5操作的k可能为负数

Source

解题思路：树状数组+线段树

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cctype>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;const int  maxn = 100005;int n, q, m;int s[maxn << 1], a[maxn], b[maxn << 1], tot;int L[maxn << 6], R[maxn << 6], sum[maxn << 6];int op[maxn], x[maxn], y[maxn], z[maxn];int lowbit(int k) { return k&-k; }void add(int &k, int l, int r, int p, int val){if (!k) k = ++tot;sum[k] += val;if (l == r) return;int mid = (l + r) / 2;if (p <= mid) add(L[k], l, mid, p, val);else add(R[k], mid + 1, r, p, val);}void update(int k, int l, int r, int p, int val){for (int i = k; i <= n; i += lowbit(i)) add(s[i], l, r, p, val);}int query1(int l, int r, int p){int A = 0, B = 0, ans = 0, a[100], b[100];for (int i = l - 1; i; i -= lowbit(i)) a[A++] = s[i];for (int i = r; i; i -= lowbit(i)) b[B++] = s[i];for (l = 1, r = m; l < r;){int cnt = 0, mid = (l + r) >> 1;for (int i = 0; i < B; i++) cnt += sum[L[b[i]]];for (int i = 0; i < A; i++) cnt -= sum[L[a[i]]];if (mid >= p){for (int i = 0; i < B; i++) b[i] = L[b[i]];for (int i = 0; i < A; i++) a[i] = L[a[i]];r = mid;}else{for (int i = 0; i < B; i++) b[i] = R[b[i]];for (int i = 0; i < A; i++) a[i] = R[a[i]];l = mid + 1, ans += cnt;}}return ans + 1;}int query2(int l, int r, int k){int A = 0, B = 0, a[100], b[100];for (int i = l - 1; i; i -= lowbit(i)) a[A++] = s[i];for (int i = r; i; i -= lowbit(i)) b[B++] = s[i];for (l = 1, r = m; l < r;){int cnt = 0, mid = (l + r) >> 1;for (int i = 0; i < B; i++) cnt += sum[L[b[i]]];for (int i = 0; i < A; i++) cnt -= sum[L[a[i]]];if (k <= cnt){for (int i = 0; i < B; i++) b[i] = L[b[i]];for (int i = 0; i < A; i++) a[i] = L[a[i]];r = mid;}else{for (int i = 0; i < B; i++) b[i] = R[b[i]];for (int i = 0; i < A; i++) a[i] = R[a[i]];l = mid + 1, k -= cnt;}}return l;}int main(){scanf("%d%d", &n, &q);for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[++m] = a[i];for (int i = 1; i <= q; i++){scanf("%d%d%d", &op[i], &x[i], &y[i]);if (op[i] != 3) scanf("%d", &z[i]);if (op[i] == 3) b[++m] = y[i];if (op[i] == 4 || op[i] == 5) b[++m] = z[i];}sort(b + 1, b + m + 1);m = unique(b + 1, b + m + 1) - b - 1;for (int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + m + 1, a[i]) - b;for (int i = 1; i <= m; i++){if (op[i] == 1 || op[i] == 4 || op[i] == 5)z[i] = lower_bound(b + 1, b + m + 1, z[i]) - b;if (op[i] == 3) y[i] = lower_bound(b + 1, b + m + 1, y[i]) - b;}for (int i = 1; i <= n; i++) update(i, 1, m, a[i], 1);for (int i = 1; i <= q; i++){if (op[i] == 1) printf("%d\n", query1(x[i], y[i], z[i]));else if (op[i] == 3){update(x[i], 1, m, a[x[i]], -1);update(x[i], 1, m, y[i], 1);a[x[i]] = y[i];}else if (op[i] == 2) printf("%d\n", b[query2(x[i], y[i], z[i])]);else if (op[i] == 4){int temp = query1(x[i], y[i], z[i]);printf("%d\n", b[query2(x[i], y[i], temp - 1)]);}else{int temp = query1(x[i], y[i], z[i] + 1);printf("%d\n", b[query2(x[i], y[i], temp)]);}}return 0;}