ZOJ 2112 Dynamic Rankings（带修改的区间第K大，分块+二分搜索+二分答案）

Posted 2020-09-22 Blackops

Dynamic Rankings

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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The Company Dynamic Rankings has developed a new kind of computer that is no longer satisfied with the query like to simply find the k-th smallest number of the given N numbers. They have developed a more powerful system such that for N numbers a[1], a[2], ..., a[N], you can ask it like: what is the k-th smallest number of a[i], a[i+1], ..., a[j]? (For some i<=j, 0<k<=j+1-i that you have given to it). More powerful, you can even change the value of some a[i], and continue to query, all the same.

Your task is to write a program for this computer, which

- Reads N numbers from the input (1 <= N <= 50,000)

- Processes M instructions of the input (1 <= M <= 10,000). These instructions include querying the k-th smallest number of a[i], a[i+1], ..., a[j] and change some a[i] to t.

Input

The first line of the input is a single number X (0 < X <= 4), the number of the test cases of the input. Then X blocks each represent a single test case.

The first line of each block contains two integers N and M, representing N numbers and M instruction. It is followed by N lines. The (i+1)-th line represents the number a[i]. Then M lines that is in the following format

Q i j k or
C i t

It represents to query the k-th number of a[i], a[i+1], ..., a[j] and change some a[i] to t, respectively. It is guaranteed that at any time of the operation. Any number a[i] is a non-negative integer that is less than 1,000,000,000.

There‘re NO breakline between two continuous test cases.

Output

For each querying operation, output one integer to represent the result. (i.e. the k-th smallest number of a[i], a[i+1],..., a[j])

There‘re NO breakline between two continuous test cases.

Sample Input

2
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3
5 3
3 2 1 4 7
Q 1 4 3
C 2 6
Q 2 5 3

Sample Output

3
6
3
6

 

 

题目链接：ZOJ 2112

突然发现用分块做很简单，每一个块维护区间内的有序序列，暴力修改+重构，查询的时候二分下答案，设当前的二分值为mid，如果区间内小于等于mid的数>=k则说明答案小于等于mid，否则答案大于mid。

代码：

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 50010;
const int M = 10010;
const int BC = 233;
struct Block
{
    int l, r;
};
Block B[M];
int arr[N], belong[N], unit, bcnt, b[N];
int n, m;

void reset(int x)
{
    for (int i = B[x].l; i <= B[x].r; ++i)
        b[i] = arr[i];
    sort(b + B[x].l, b + B[x].r + 1);
}
void init()
{
    unit = sqrt(n);
    bcnt = n / unit;
    if (n % unit)
        ++bcnt;
    int i;
    for (i = 1; i <= bcnt; ++i)
    {
        B[i].l = (i - 1) * unit + 1;
        B[i].r = i * unit;
    }
    B[bcnt].r = n;
    for (i = 1; i <= n; ++i)
        belong[i] = (i - 1) / unit + 1;
    for (i = 1; i <= bcnt; ++i)
        reset(i);
}
void update(int x, int t)
{
    int bx = belong[x];
    arr[x] = t;
    reset(bx);
}
int bs(int x, int key)
{
    int l = B[x].l, r = B[x].r;
    int ans = -1;
    while (l <= r)
    {
        int mid = MID(l, r);
        if (b[mid] <= key)
        {
            ans = mid;
            l = mid + 1;
        }
        else
            r = mid - 1;
    }
    return ~ans ? ans - B[x].l + 1 : 0;
}
int query(int l, int r, int k)
{
    int L = 0, R = 1e9;
    int ans = 1;
    int bl = belong[l], br = belong[r], i;
    while (L <= R)
    {
        int tk = 0;
        int mid = MID(L, R);
        for (i = l; i <= B[bl].r; ++i)
            if (arr[i] <= mid)
                ++tk;
        for (i = B[br].l; i <= r; ++i)
            if (arr[i] <= mid)
                ++tk;
        for (i = bl + 1; i < br; ++i)
            tk += bs(i, mid);
        if (tk >= k)
        {
            ans = mid;
            R = mid - 1;
        }
        else
            L = mid + 1;
    }
    return ans;
}
int main(void)
{
    int tcase, i;
    char ops[3];
    int l, r, k, x, t;
    scanf("%d", &tcase);
    while (tcase--)
    {
        scanf("%d%d", &n, &m);
        for (i = 1; i <= n; ++i)
            scanf("%d", &arr[i]);
        init();
        for (i = 1; i <= m; ++i)
        {
            scanf("%s", ops);
            if (ops[0] == ‘Q‘)
            {
                scanf("%d%d%d", &l, &r, &k);
                printf("%d\n", query(l, r, k));
            }
            else
            {
                scanf("%d%d", &x, &t);
                update(x, t);
            }
        }
    }
    return 0;
}