Luogu P2574 XOR的艺术 (线段树)
Problem
题目描述
AKN觉得第一题太水了,不屑于写第一题,所以他又玩起了新的游戏。在游戏中,他发现,这个游戏的伤害计算有一个规律,规律如下
- 拥有一个伤害串为长度为\(n\)的\(01\)串。
- 给定一个范围\([l,r]\),伤害为伤害串的这个范围内中\(1\)的个数
- 会被随机修改伤害串中的数值,修改的方法是把\([l,r]\)中的所有数\(\oplus\)上\(1\)
AKN想知道一些时刻的伤害,请你帮助他求出这个伤害
输入输出格式
输入格式:
第一行两个数\(n\),\(m\),表示长度为\(n\)的\(01\)串,有\(m\)个时刻
第二行一个长度为\(n\)的\(01\)串,为初始伤害串
第三行开始\(m\)行,每行三个数\(p\),\(l\),\(r\)
若\(p\)为\(0\),则表示当前时刻改变\([l,r]\)的伤害串,改变规则如上
若\(p\)为\(1\),则表示当前时刻AKN想知道\([l,r]\)的伤害
输出格式:
对于每次询问伤害,输出一个数值伤害,每次询问输出一行
输入输出样例
输入样例#1:
10 6
1011101001
0 2 4
1 1 5
0 3 7
1 1 10
0 1 4
1 2 6
输出样例#1:
3
6
1
说明
样例解释:
\(1011101001\)
\(1100101001\)
询问\([1,5]\)输出\(3\)
\(1111010001\)
询问\([1,10]\)输出\(6\)
\(0000010001\)
询问\([2,6]\)输出\(1\)
数据范围:
\(10\%\)数据\(2\leq n\),\(m\leq 10\)
另有\(30\%\)数据\(2\leq n,m\leq 2000\)
\(100\%\)数据\(2\leq n,m\leq 2\times 10^5\)
By:worcher
Solution
好像线段树写上瘾了
挺简单的一道线段树, 异或操作交换\(0\)和\(1\)的数量即可。
#include <algorithm>
#include <cstdio>
class SegmentTree {
private:
struct Node {
int zero, one, delta;
Node *left, *right;
int L, R;
inline void Update() {
if (this->left) {
this->zero = this->left->zero + this->right->zero;
this->one = this->left->one + this->right->one;
}
}
inline void PushDown() {
if (this->delta && this->left) {
std::swap(this->left->zero, this->left->one);
std::swap(this->right->zero, this->right->one);
this->left->delta ^= 1;
this->right->delta ^= 1;
this->delta = 0;
}
}
Node(const int &l, const int &r) {
this->L = l, this->R = r;
this->zero = this->one = this->delta = 0;
this->left = this->right = nullptr;
if (l == r) {
if (getchar() - '0')
this->one = 1;
else
this->zero = 1;
} else {
register int mid((l + r) >> 1);
this->left = new Node(l, mid);
this->right = new Node(mid + 1, r);
this->Update();
}
}
~Node() {
if (this->left) {
delete this->left;
delete this->right;
}
}
inline int Sum(const int &l, const int &r) {
if (l <= this->L && this->R <= r) {
return this->one;
} else {
this->PushDown();
register int mid((this->L + this->R) >> 1), ret(0);
if (l <= mid) ret += this->left->Sum(l, r);
if (mid < r) ret += this->right->Sum(l, r);
return ret;
}
}
inline void ExclusiveOr(const int &l, const int &r) {
if (l <= this->L && this->R <= r) {
std::swap(this->zero, this->one);
this->delta ^= 1;
} else {
this->PushDown();
register int mid((this->L + this->R) >> 1);
if (l <= mid) this->left->ExclusiveOr(l, r);
if (mid < r) this->right->ExclusiveOr(l, r);
this->Update();
}
}
} * root;
public:
SegmentTree() { this->root = nullptr; }
inline void Clear() {
if (this->root) {
delete this->root;
this->root = nullptr;
}
}
inline void Build(const int &l, const int &r) { this->root = new Node(l, r); }
inline int Sum(const int &l, const int &r) { return this->root->Sum(l, r); }
inline void ExclusiveOr(const int &l, const int &r) {
this->root->ExclusiveOr(l, r);
}
} T;
int n, m;
int p, l, r;
int main(int argc, char const *argv[]) {
scanf("%d %d\n", &n, &m);
T.Build(1, n);
while (m--) {
scanf("%d %d %d", &p, &l, &r);
switch (p) {
case 0: {
T.ExclusiveOr(l, r);
break;
}
case 1: {
printf("%d\n", T.Sum(l, r));
break;
}
}
}
return 0;
}