【线段树】P9349 [JOI 2023 Final] Stone Arranging 2|普及+

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C++线段树

P9349 [JOI 2023 Final] Stone Arranging 2

题目描述

JOI-kun has $N$ go stones. The stones are numbered from $1$ to $N$. The color of each stone is an integer between $1$ and $10^9$, inclusive. In the beginning, the color of Stone $i$ ($1\le i\le N$) is $A_i$.

From now, JOI-kun will perform $N$ operations. He will put the stones on the table in a line. The operation $i$ ($1\le i\le N$) will be performed as follows:

1. JOI-kun will put Stone $i$ on the immediate right of Stone $i-1$. However, when $i=1$, JOI-kun will put Stone 1 on the table.
2. If there is a stone among Stones $1,2,\cdots ,i-1$ whose current color is the same as Stone $i$, let $j$ be the maximum index of such stones, and JOI-kun will paint all of Stones $j+1,j+2,\cdots ,i-1$ with the color $A_i$.

In order to confirm whether the operations are correctly performed, JOI-kun wants to know in advance the colors of the stones after all the operations are performed.

Given information of the go stones, write a program which determines the colors of the stones after the $N$ operations are performed.

输入格式

Read the following data from the standard input.

$N$
$A_1$
$A_2$
$⋮$
$A_N$

输出格式

Write $N$ lines to the standard output. The $i$-th line ($1\le i\le N$) should contain the color of Stone $i$ after the $N$ operations are performed.

输入输出样例 #1

输入 #1

6
1
2
1
2
3
2

输出 #1

1
1
1
2
2
2

输入输出样例 #2

输入 #2

10
1
1
2
2
1
2
2
1
1
2

输出 #2

1
1
1
1
1
1
1
1
1
2

说明/提示

Samples

Sample 1

The operations are performed as in the following table.

Finally, the colors of Stones 1, 2, 3, 4, 5, 6 will be 1, 1, 1, 2, 2, 2, respectively.

This sample input satisfies the constraints of Subtasks 1, 3.

Sample 2

This sample input satisfies the constraints of all the subtasks.

Constraints

• $1\le N\le 2\times 10^5$.
• $1\le A_i\le 10^9$ ($1\le i\le N$).
• Given values are all integers.

Subtasks

1. (25 points) $N\le 2000$.
2. (35 points) $A_i\le 2$ ($1\le i\le N$).
3. (40 points) No additional constraints.

线段树

以下两个变量记录所有已经排列，且不属于任意a[j…i-1]的棋子。
unordered_map<int, vector> mColorIndexs; 键是颜色，值是下标，升序。
set inxs;下标。
线段树记录修改状态。

代码

核心代码

#include <iostream>
#include <sstream>
#include <vector>
#include<map>
#include<unordered_map>
#include<set>
#include<unordered_set>
#include<string>
#include<algorithm>
#include<functional>
#include<queue>
#include <stack>
#include<iomanip>
#include<numeric>
#include <math.h>
#include <climits>
#include<assert.h>
#include<cstring>
#include<list>#include <bitset>
using namespace std;template<class T1, class T2>
std::istream& operator >> (std::istream& in, pair<T1, T2>& pr) {in >> pr.first >> pr.second;return in;
}template<class T1, class T2, class T3 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3>& t) {in >> get<0>(t) >> get<1>(t) >> get<2>(t);return in;
}template<class T1, class T2, class T3, class T4 >
std::istream& operator >> (std::istream& in, tuple<T1, T2, T3, T4>& t) {in >> get<0>(t) >> get<1>(t) >> get<2>(t) >> get<3>(t);return in;
}template<class T = int>
vector<T> Read() {int n;scanf("%d", &n);vector<T> ret(n);for (int i = 0; i < n; i++) {cin >> ret[i];}return ret;
}template<class T = int>
vector<T> Read(int n) {vector<T> ret(n);for (int i = 0; i < n; i++) {cin >> ret[i];}return ret;
}template<int N = 12 * 1'000'000>
class COutBuff
{
public:COutBuff() {m_p = puffer;}template<class T>void write(T x) {int num[28], sp = 0;if (x < 0)*m_p++ = '-', x = -x;if (!x)*m_p++ = 48;while (x)num[++sp] = x % 10, x /= 10;while (sp)*m_p++ = num[sp--] + 48;}inline void write(char ch){*m_p++ = ch;}inline void ToFile() {fwrite(puffer, 1, m_p - puffer, stdout);}
private:char  puffer[N], * m_p;
};template<int N = 12 * 1'000'000>
class CInBuff
{
public:inline CInBuff() {fread(buffer, 1, N, stdin);}inline int Read() {int x(0), f(0);while (!isdigit(*S))f |= (*S++ == '-');while (isdigit(*S))x = (x << 1) + (x << 3) + (*S++ ^ 48);return f ? -x : x;}
private:char buffer[N], * S = buffer;
};template<class TSave, class TRecord >
class CRangUpdateLineTree
{
protected:virtual void OnQuery(TSave& ans, const TSave& save, const int& iSaveLeft, const int& iSaveRight) = 0;virtual void OnUpdate(TSave& save, const int& iSaveLeft, const int& iSaveRight, const TRecord& update) = 0;virtual void OnUpdateParent(TSave& par, const TSave& left, const TSave& r, const int& iSaveLeft, const int& iSaveRight) = 0;virtual void OnUpdateRecord(TRecord& old, const TRecord& newRecord) = 0;
};template<class TSave, class TRecord >
class CVectorRangeUpdateLineTree : public CRangUpdateLineTree<TSave, TRecord>
{
public:CVectorRangeUpdateLineTree(int iEleSize, TSave tDefault, TRecord tRecordNull) :m_iEleSize(iEleSize), m_tDefault(tDefault), m_save(iEleSize * 4, tDefault), m_record(iEleSize * 4, tRecordNull) {m_recordNull = tRecordNull;}void Update(int iLeftIndex, int iRightIndex, TRecord value){Update(1, 0, m_iEleSize - 1, iLeftIndex, iRightIndex, value);}TSave Query(int leftIndex, int rightIndex) {return Query(leftIndex, rightIndex, m_tDefault);}TSave Query(int leftIndex, int rightIndex, const TSave& tDefault) {TSave ans = tDefault;Query(ans, 1, 0, m_iEleSize - 1, leftIndex, rightIndex);return ans;}//void Init() {//	Init(1, 0, m_iEleSize - 1);//}TSave QueryAll() {return m_save[1];}void swap(CVectorRangeUpdateLineTree<TSave, TRecord>& other) {m_save.swap(other.m_save);m_record.swap(other.m_record);std::swap(m_recordNull, other.m_recordNull);assert(m_iEleSize == other.m_iEleSize);}
protected://void Init(int iNodeNO, int iSaveLeft, int iSaveRight)//{//	if (iSaveLeft == iSaveRight) {//		this->OnInit(m_save[iNodeNO], iSaveLeft);//		return;//	}//	const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;//	Init(iNodeNO * 2, iSaveLeft, mid);//	Init(iNodeNO * 2 + 1, mid + 1, iSaveRight);//	this->OnUpdateParent(m_save[iNodeNO], m_save[iNodeNO * 2], m_save[iNodeNO * 2 + 1], iSaveLeft, iSaveRight);//}void Query(TSave& ans, int iNodeNO, int iSaveLeft, int iSaveRight, int iQueryLeft, int iQueryRight) {if ((iSaveLeft >= iQueryLeft) && (iSaveRight <= iQueryRight)) {this->OnQuery(ans, m_save[iNodeNO], iSaveLeft, iSaveRight);return;}if (iSaveLeft == iSaveRight) {//没有子节点return;}Fresh(iNodeNO, iSaveLeft, iSaveRight);const int mid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;if (mid >= iQueryLeft) {Query(ans, iNodeNO * 2, iSaveLeft, mid, iQueryLeft, iQueryRight);}if (mid + 1 <= iQueryRight) {Query(ans, iNodeNO * 2 + 1, mid + 1, iSaveRight, iQueryLeft, iQueryRight);}}void Update(int iNode, int iSaveLeft, int iSaveRight, int iOpeLeft, int iOpeRight, TRecord value){if ((iOpeLeft <= iSaveLeft) && (iOpeRight >= iSaveRight)){this->OnUpdate(m_save[iNode], iSaveLeft, iSaveRight, value);this->OnUpdateRecord(m_record[iNode], value);return;}if (iSaveLeft == iSaveRight) {return;//没有子节点}Fresh(iNode, iSaveLeft, iSaveRight);const int iMid = iSaveLeft + (iSaveRight - iSaveLeft) / 2;if (iMid >= iOpeLeft){Update(iNode * 2, iSaveLeft, iMid, iOpeLeft, iOpeRight, value);}if (iMid + 1 <= iOpeRight){Update(iNode * 2 + 1, iMid + 1, iSaveRight, iOpeLeft, iOpeRight, value);}// 如果有后代，至少两个后代this->OnUpdateParent(m_save[iNode], m_save[iNode * 2], m_save[iNode * 2 + 1], iSaveLeft, iSaveRight);}void Fresh(int iNode, int iDataLeft, int iDataRight){if (m_recordNull == m_record[iNode]){return;}const int iMid = iDataLeft + (iDataRight - iDataLeft) / 2;Update(iNode * 2, iDataLeft, iMid, iDataLeft, iMid, m_record[iNode]);Update(iNode * 2 + 1, iMid + 1, iDataRight, iMid + 1, iDataRight, m_record[iNode]);m_record[iNode] = m_recordNull;}vector<TSave> m_save;vector<TRecord> m_record;TRecord m_recordNull;TSave m_tDefault;const int m_iEleSize;
};template<class TSave, class TRecord >
class  CSetMaxLineTree : public CVectorRangeUpdateLineTree<TSave, TRecord>
{
public:using CVectorRangeUpdateLineTree<TSave, TRecord>::CVectorRangeUpdateLineTree;
protected:virtual void OnQuery(TSave& ans, const TSave& save, const int& iSaveLeft, const int& iSaveRight) {ans = max(ans, save);}virtual void OnUpdate(TSave& save, const int& iSaveLeft, const int& iSaveRight, const TRecord& update) {save = update;}virtual void OnUpdateParent(TSave& par, const TSave& left, const TSave& r, const int& iSaveLeft, const int& iSaveRight) {par = max(left, r);}virtual void OnUpdateRecord(TRecord& old, const TRecord& newRecord){old = newRecord;}
};class Solution {
public:vector<int> Ans(const vector<int>& a) {const int N = a.size();CSetMaxLineTree<int, int> lineTree(N, 0, 0);set<int> inxs;unordered_map<int, vector<int>> mColorIndexs;for (int i = 0; i < N; i++) {int j = i;if (mColorIndexs.count(a[i]) && mColorIndexs[a[i]].size()) {j = mColorIndexs[a[i]].back();};for (auto it = inxs.lower_bound(i);;) {if (inxs.begin() == it) { break; }--it;if (*it < j) { break; }mColorIndexs[a[*it]].pop_back();}inxs.erase(inxs.lower_bound(j), inxs.lower_bound(i));inxs.emplace(i);mColorIndexs[a[i]].emplace_back(i);lineTree.Update(j, i, a[i]);}vector<int> ans;for (int i = 0; i < N; i++) {ans.emplace_back(lineTree.Query(i, i));}return ans;}
};int main() {	
#ifdef _DEBUGfreopen("a.in", "r", stdin);
#endif // DEBUG	auto a = Read<int>();auto res = Solution().Ans(a);for (const auto& i : res) {cout << i << endl;}cout << endl;
#ifdef _DEBUG		/*printf("T=%d,", T);*/Out(a, "a=");
#endif // DEBUG	return 0;
}

单元测试

vector<int> a;TEST_METHOD(TestMethod11){a = {1,2};auto res = Solution().Ans(a);AssertV({1,2}, res);}TEST_METHOD(TestMethod12){a = { 1,2 ,1};auto res = Solution().Ans(a);AssertV({ 1,1,1 }, res);}TEST_METHOD(TestMethod13){a = { 4,1,2,3,2,1 };auto res = Solution().Ans(a);AssertV({4,1,1, 1,1,1 }, res);}TEST_METHOD(TestMethod14){a = { 1,2,1,2,3,2 };auto res = Solution().Ans(a);AssertV({ 1,1,1,2,2,2 }, res);}TEST_METHOD(TestMethod15){a = { 1,1,2,2,1,2,2,1,1,2 };auto res = Solution().Ans(a);AssertV({ 1,1,1,1,1,1,1,1,1,2 }, res);}

不用线段树

在inxs中存在的下标，颜色是原色，不存在的颜色就是inxs存在的后一个下标的颜色。
即：a[i] = *inxs.lower(i);

class Solution {public:vector<int> Ans(const vector<int>& a) {const int N = a.size();set<int> inxs;unordered_map<int, vector<int>> mColorIndexs;for (int i = 0; i < N; i++) {int j = i;if (mColorIndexs.count(a[i]) && mColorIndexs[a[i]].size()) {j = mColorIndexs[a[i]].back();};for (auto it = inxs.lower_bound(i);;) {if (inxs.begin() == it) { break; }--it;if (*it < j) { break; }mColorIndexs[a[*it]].pop_back();}inxs.erase(inxs.lower_bound(j), inxs.lower_bound(i));inxs.emplace(i);mColorIndexs[a[i]].emplace_back(i);					;}vector<int> ans;for (int i = 0; i < N; i++) {ans.emplace_back(a[*inxs.lower_bound(i)]);}return ans;}};

扩展阅读

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如果程序是一条龙，那算法就是他的是睛
失败+反思=成功 成功+反思=成功

视频课程

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测试环境

操作系统：win7 开发环境： VS2019 C++17
或者 操作系统：win10 开发环境： VS2022 C++17
如无特殊说明，本算法用**C++**实现。