在二进制数组中将范围更新查询与1进行XOR。

给定大小为N的二进制数组arr [] 。任务是回答Q查询，该查询可以是以下任意一种：
类型1 – 1 lr：使用1对从l到r的所有数组元素执行按位异或运算。
类型2 – 2 lr：返回子数组[l，r]中值为1的两个元素之间的最小距离。
类型3 – 3 lr：返回子数组[l，r]中值为1的两个元素之间的最大距离。
类型4 – 4 lr：返回子数组[l，r]中值为0的两个元素之间的最小距离。
类型5 – 5 lr：返回子数组[l，r]中值为0的两个元素之间的最大距离。
例子：

Input : arr[] = {1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0}, q=5 
Output : 2 2 3 2
Explanation :
query 1 : Type 2, l=3, r=7 
Range 3 to 7 contains { 1, 0, 1, 0, 1 }. 
So, the minimum distance between two elements with value 1 is 2.
query 2 : Type 3, l=2, r=5 
Range 2 to 5 contains { 0, 1, 0, 1 }. 
So, the maximum distance between two elements with value 1 is 2.
query 3 : Type 1, l=1, r=4 
After update array becomes {1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0}
query 4 : Type 4, l=3, r=7 
Range 3 to 7 in updated array contains { 0, 1, 1, 0, 1 }. 
So, the minimum distance between two elements with value 0 is 3.
query 5 : Type 5, l=4, r=9 
Range 4 to 9 contains { 1, 1, 0, 1, 0, 1 }. 
So, the maximum distance between two elements with value 0 is 2.
 

方法：
我们将创建一个细分树，并使用具有延迟传播的范围更新来解决此问题。

1. 段树中的每个节点将具有最左1以及最右1，最左0以及最右0的索引，以及包含子数组{l，r}中任何值为1的元素之间的最大和最小距离的整数作为子数组{l，r}中值为0的任何元素之间的最大和最小距离。
2. 现在，在此段树中，我们可以按如下所示合并左右节点：

CPP

// l1 = leftmost index of 1, l0 = leftmost index of 0.
// r1 = rightmost index of 1, r0 = rightmost index of 0.
// max1 = maximum distance between two 1’s.
// max0 = maximum distance between two 0’s.
// min1 = minimum distance between two 1’s.
// min0 = minimum distance between two 0’s.
node Merge(node left, node right)
{
    node cur;
     
    if left.l0 is valid
        cur.l0 = left.l0
    else
        cur.l0 = r.l0
    // We will do this for all values
    // i.e. cur.r0, cur.l1, cur.r1, cur.l0
     
    // To find the min and max difference between two 1's and 0's
    // we will take min/max value of left side, right side and
    // difference between rightmost index of 1/0 in right node
    // and leftmost index of 1/0 in left node respectively.
         
     cur.min0 = minimum of left.min0 and right.min0
  
     if left.r0 is valid and right.l0 is valid
        cur.min0 = minimum of cur.min0 and (right.l0 - left.r0)
    // We will do this for all max/min values
    // i.e. cur.min0, cur.min1, cur.max1, cur.max0
         
    return cur;
}

2. 为了处理范围更新查询，我们将使用延迟传播。更新查询要求我们将l到r范围内的所有元素与1进行异或运算，从观察值中我们知道：

0 xor 1 = 1
       1 xor 1 = 0

1. 因此，我们可以观察到，在此更新之后，所有0都将变为1，而所有1都将变为0。因此，在我们的段树节点中，0和1的所有对应值也将被交换，即

l0 and l1 will get swapped
       r0 and r1 will get swapped
       min0 and min1 will get swapped
       max0 and max1 will get swapped

2. 然后，最后要找到任务2、3、4和5的答案，我们只需要调用给定范围{l，r}的查询函数，而我要找到任务1的答案，我们需要调用范围更新函数。

下面是上述方法的实现：

CPP

// C++ program for the given problem
#include 
using namespace std;
 
int lazy[100001];
 
// Class for each node
// in the segment tree
class node {
public:
    int l1, r1, l0, r0;
    int min0, max0, min1, max1;
 
    node()
    {
        l1 = r1 = l0 = r0 = -1;
 
        max1 = max0 = INT_MIN;
        min1 = min0 = INT_MAX;
    }
 
} seg[100001];
 
// A utility function for
// merging two nodes
node MergeUtil(node l, node r)
{
    node x;
 
    x.l0 = (l.l0 != -1) ? l.l0 : r.l0;
    x.r0 = (r.r0 != -1) ? r.r0 : l.r0;
 
    x.l1 = (l.l1 != -1) ? l.l1 : r.l1;
    x.r1 = (r.r1 != -1) ? r.r1 : l.r1;
 
    x.min0 = min(l.min0, r.min0);
    if (l.r0 != -1 && r.l0 != -1)
        x.min0 = min(x.min0, r.l0 - l.r0);
 
    x.min1 = min(l.min1, r.min1);
    if (l.r1 != -1 && r.l1 != -1)
        x.min1 = min(x.min1, r.l1 - l.r1);
 
    x.max0 = max(l.max0, r.max0);
    if (l.l0 != -1 && r.r0 != -1)
        x.max0 = max(x.max0, r.r0 - l.l0);
 
    x.max1 = max(l.max1, r.max1);
    if (l.l1 != -1 && r.r1 != -1)
        x.max1 = max(x.max1, r.r1 - l.l1);
 
    return x;
}
 
// utility function
// for updating a node
node UpdateUtil(node x)
{
    swap(x.l0, x.l1);
    swap(x.r0, x.r1);
    swap(x.min1, x.min0);
    swap(x.max0, x.max1);
 
    return x;
}
 
// A recursive function that constructs
// Segment Tree for given string
void Build(int qs, int qe, int ind, int arr[])
{
    // If start is equal to end then
    // insert the array element
    if (qs == qe) {
        if (arr[qs] == 1) {
            seg[ind].l1 = seg[ind].r1 = qs;
        }
        else {
            seg[ind].l0 = seg[ind].r0 = qs;
        }
 
        lazy[ind] = 0;
        return;
    }
    int mid = (qs + qe) >> 1;
 
    // Build the segment tree
    // for range qs to mid
    Build(qs, mid, ind << 1, arr);
 
    // Build the segment tree
    // for range mid+1 to qe
    Build(mid + 1, qe, ind << 1 | 1, arr);
 
    // merge the two child nodes
    // to obtain the parent node
    seg[ind] = MergeUtil(
        seg[ind << 1],
        seg[ind << 1 | 1]);
}
 
// Query in a range qs to qe
node Query(int qs, int qe,
           int ns, int ne, int ind)
{
    if (lazy[ind] != 0) {
        seg[ind] = UpdateUtil(seg[ind]);
        if (ns != ne) {
            lazy[ind * 2] ^= lazy[ind];
            lazy[ind * 2 + 1] ^= lazy[ind];
        }
        lazy[ind] = 0;
    }
 
    node x;
 
    // If the range lies in this segment
    if (qs <= ns && qe >= ne)
        return seg[ind];
 
    // If the range is out of the bounds
    // of this segment
    if (ne < qs || ns > qe || ns > ne)
        return x;
 
    // Else query for the right and left
    // child node of this subtree
    // and merge them
    int mid = (ns + ne) >> 1;
 
    node l = Query(qs, qe, ns,
                   mid, ind << 1);
    node r = Query(qs, qe,
                   mid + 1, ne,
                   ind << 1 | 1);
 
    x = MergeUtil(l, r);
    return x;
}
 
// range update using lazy prpagation
void RangeUpdate(int us, int ue,
                 int ns, int ne, int ind)
{
    if (lazy[ind] != 0) {
        seg[ind] = UpdateUtil(seg[ind]);
        if (ns != ne) {
            lazy[ind * 2] ^= lazy[ind];
            lazy[ind * 2 + 1] ^= lazy[ind];
        }
        lazy[ind] = 0;
    }
 
    // If the range is out of the bounds
    // of this segment
    if (ns > ne || ns > ue || ne < us)
        return;
 
    // If the range lies in this segment
    if (ns >= us && ne <= ue) {
        seg[ind] = UpdateUtil(seg[ind]);
        if (ns != ne) {
            lazy[ind * 2] ^= 1;
            lazy[ind * 2 + 1] ^= 1;
        }
        return;
    }
 
    // Else query for the right and left
    // child node of this subtree
    // and merge them
    int mid = (ns + ne) >> 1;
    RangeUpdate(us, ue, ns, mid, ind << 1);
    RangeUpdate(us, ue, mid + 1, ne, ind << 1 | 1);
 
    node l = seg[ind << 1], r = seg[ind << 1 | 1];
    seg[ind] = MergeUtil(l, r);
}
 
// Driver code
int main()
{
 
    int arr[] = { 1, 1, 0,
                  1, 0, 1,
                  0, 1, 0,
                  1, 0, 1,
                  1, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Build the segment tree
    Build(0, n - 1, 1, arr);
 
    // Query of Type 2 in the range 3 to 7
    node ans = Query(3, 7, 0, n - 1, 1);
    cout << ans.min1 << "\n";
 
    // Query of Type 3 in the range 2 to 5
    ans = Query(2, 5, 0, n - 1, 1);
    cout << ans.max1 << "\n";
 
    // Query of Type 1 in the range 1 to 4
    RangeUpdate(1, 4, 0, n - 1, 1);
 
    // Query of Type 4 in the range 3 to 7
    ans = Query(3, 7, 0, n - 1, 1);
    cout << ans.min0 << "\n";
 
    // Query of Type 5 in the range 4 to 9
    ans = Query(4, 9, 0, n - 1, 1);
    cout << ans.max0 << "\n";
 
    return 0;
}

Python3

# Python program for the given problem
from sys import maxsize
from typing import List
INT_MAX = maxsize
INT_MIN = -maxsize
lazy = [0 for _ in range(100001)]
 
# Class for each node
# in the segment tree
class node:
    def __init__(self) -> None:
        self.l1 = self.r1 = self.l0 = self.r0 = -1
        self.max0 = self.max1 = INT_MIN
        self.min0 = self.min1 = INT_MAX
 
seg = [node() for _ in range(100001)]
 
# A utility function for
# merging two nodes
def MergeUtil(l: node, r: node) -> node:
    x = node()
 
    x.l0 = l.l0 if (l.l0 != -1) else r.l0
    x.r0 = r.r0 if (r.r0 != -1) else l.r0
 
    x.l1 = l.l1 if (l.l1 != -1) else r.l1
    x.r1 = r.r1 if (r.r1 != -1) else l.r1
 
    x.min0 = min(l.min0, r.min0)
    if (l.r0 != -1 and r.l0 != -1):
        x.min0 = min(x.min0, r.l0 - l.r0)
 
    x.min1 = min(l.min1, r.min1)
    if (l.r1 != -1 and r.l1 != -1):
        x.min1 = min(x.min1, r.l1 - l.r1)
 
    x.max0 = max(l.max0, r.max0)
    if (l.l0 != -1 and r.r0 != -1):
        x.max0 = max(x.max0, r.r0 - l.l0)
 
    x.max1 = max(l.max1, r.max1)
    if (l.l1 != -1 and r.r1 != -1):
        x.max1 = max(x.max1, r.r1 - l.l1)
 
    return x
 
# utility function
# for updating a node
def UpdateUtil(x: node) -> node:
    x.l0, x.l1 = x.l1, x.l0
    x.r0, x.r1 = x.r1, x.r0
    x.min1, x.min0 = x.min0, x.min1
    x.max0, x.max1 = x.max1, x.max0
 
    return x
 
# A recursive function that constructs
# Segment Tree for given string
def Build(qs: int, qe: int, ind: int, arr: List[int]) -> None:
 
  # If start is equal to end then
    # insert the array element
    if (qs == qe):
        if (arr[qs] == 1):
            seg[ind].l1 = seg[ind].r1 = qs
        else:
            seg[ind].l0 = seg[ind].r0 = qs
 
        lazy[ind] = 0
        return
 
    mid = (qs + qe) >> 1
 
    # Build the segment tree
    # for range qs to mid
    Build(qs, mid, ind << 1, arr)
 
    # Build the segment tree
    # for range mid+1 to qe
    Build(mid + 1, qe, ind << 1 | 1, arr)
 
    # merge the two child nodes
    # to obtain the parent node
    seg[ind] = MergeUtil(seg[ind << 1], seg[ind << 1 | 1])
 
# Query in a range qs to qe
def Query(qs: int, qe: int, ns: int, ne: int, ind: int) -> node:
    if (lazy[ind] != 0):
        seg[ind] = UpdateUtil(seg[ind])
        if (ns != ne):
            lazy[ind * 2] ^= lazy[ind]
            lazy[ind * 2 + 1] ^= lazy[ind]
        lazy[ind] = 0
    x = node()
 
    # If the range lies in this segment
    if (qs <= ns and qe >= ne):
        return seg[ind]
 
    # If the range is out of the bounds
    # of this segment
    if (ne < qs or ns > qe or ns > ne):
        return x
 
    # Else query for the right and left
    # child node of this subtree
    # and merge them
    mid = (ns + ne) >> 1
    l = Query(qs, qe, ns, mid, ind << 1)
    r = Query(qs, qe, mid + 1, ne, ind << 1 | 1)
    x = MergeUtil(l, r)
    return x
 
# range update using lazy prpagation
def RangeUpdate(us: int, ue: int, ns: int, ne: int, ind: int) -> None:
    if (lazy[ind] != 0):
        seg[ind] = UpdateUtil(seg[ind])
        if (ns != ne):
            lazy[ind * 2] ^= lazy[ind]
            lazy[ind * 2 + 1] ^= lazy[ind]
        lazy[ind] = 0
 
    # If the range is out of the bounds
    # of this segment
    if (ns > ne or ns > ue or ne < us):
        return
 
    # If the range lies in this segment
    if (ns >= us and ne <= ue):
        seg[ind] = UpdateUtil(seg[ind])
        if (ns != ne):
            lazy[ind * 2] ^= 1
            lazy[ind * 2 + 1] ^= 1
        return
 
    # Else query for the right and left
    # child node of this subtree
    # and merge them
    mid = (ns + ne) >> 1
    RangeUpdate(us, ue, ns, mid, ind << 1)
    RangeUpdate(us, ue, mid + 1, ne, ind << 1 | 1)
    l = seg[ind << 1]
    r = seg[ind << 1 | 1]
    seg[ind] = MergeUtil(l, r)
 
# Driver code
if __name__ == "__main__":
    arr = [1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0]
    n = len(arr)
 
    # Build the segment tree
    Build(0, n - 1, 1, arr)
 
    # Query of Type 2 in the range 3 to 7
    ans = Query(3, 7, 0, n - 1, 1)
    print(ans.min1)
 
    # Query of Type 3 in the range 2 to 5
    ans = Query(2, 5, 0, n - 1, 1)
    print(ans.max1)
 
    # Query of Type 1 in the range 1 to 4
    RangeUpdate(1, 4, 0, n - 1, 1)
 
    # Query of Type 4 in the range 3 to 7
    ans = Query(3, 7, 0, n - 1, 1)
    print(ans.min0)
 
    # Query of Type 5 in the range 4 to 9
    ans = Query(4, 9, 0, n - 1, 1)
    print(ans.max0)
 
# This code is contributed by sanjeev2552

输出：

2
2
3
2