当前位置：首页 > news >正文

辽宁建设工程信息网业绩录入/百度优化软件

news 文章来源：https://blog.csdn.net/he_zhidan/article/details/135242456 2025/3/11 3:22:34

辽宁建设工程信息网业绩录入,百度优化软件,apicloud安装wordpress,做旅游的网站涉及知识点动态规划多源最短路径字典树题目给你两个下标从 0 开始的字符串 source 和 target ，它们的长度均为 n 并且由小写英文字母组成。另给你两个下标从 0 开始的字符串数组 original 和 changed ，以及一个整数数组 cost ，其中…

涉及知识点

动态规划多源最短路径字典树

题目

给你两个下标从 0 开始的字符串 source 和 target ，它们的长度均为 n 并且由小写英文字母组成。
另给你两个下标从 0 开始的字符串数组 original 和 changed ，以及一个整数数组 cost ，其中 cost[i] 代表将字符串 original[i] 更改为字符串 changed[i] 的成本。
你从字符串 source 开始。在一次操作中，如果存在任意下标 j 满足 cost[j] == z 、original[j] == x 以及 changed[j] == y ，你就可以选择字符串中的子串 x 并以 z 的成本将其更改为 y 。你可以执行任意数量的操作，但是任两次操作必须满足以下两个条件之一：
在两次操作中选择的子串分别是 source[a…b] 和 source[c…d] ，满足 b < c 或 d < a 。换句话说，两次操作中选择的下标不相交。
在两次操作中选择的子串分别是 source[a…b] 和 source[c…d] ，满足 a == c 且 b == d 。换句话说，两次操作中选择的下标相同。
返回将字符串 source 转换为字符串 target 所需的最小成本。如果不可能完成转换，则返回 -1 。
注意，可能存在下标 i 、j 使得 original[j] == original[i] 且 changed[j] == changed[i] 。
示例 1：
输入：source = “abcd”, target = “acbe”, original = [“a”,“b”,“c”,“c”,“e”,“d”], changed = [“b”,“c”,“b”,“e”,“b”,“e”], cost = [2,5,5,1,2,20]
输出：28
解释：将 “abcd” 转换为 “acbe”，执行以下操作：

将子串 source[1…1] 从 “b” 改为 “c” ，成本为 5 。
将子串 source[2…2] 从 “c” 改为 “e” ，成本为 1 。
将子串 source[2…2] 从 “e” 改为 “b” ，成本为 2 。
将子串 source[3…3] 从 “d” 改为 “e” ，成本为 20 。
产生的总成本是 5 + 1 + 2 + 20 = 28 。
可以证明这是可能的最小成本。
示例 2：
输入：source = “abcdefgh”, target = “acdeeghh”, original = [“bcd”,“fgh”,“thh”], changed = [“cde”,“thh”,“ghh”], cost = [1,3,5]
输出：9
解释：将 “abcdefgh” 转换为 “acdeeghh”，执行以下操作：
将子串 source[1…3] 从 “bcd” 改为 “cde” ，成本为 1 。
将子串 source[5…7] 从 “fgh” 改为 “thh” ，成本为 3 。可以执行此操作，因为下标 [5,7] 与第一次操作选中的下标不相交。
将子串 source[5…7] 从 “thh” 改为 “ghh” ，成本为 5 。可以执行此操作，因为下标 [5,7] 与第一次操作选中的下标不相交，且与第二次操作选中的下标相同。
产生的总成本是 1 + 3 + 5 = 9 。
可以证明这是可能的最小成本。
示例 3：
输入：source = “abcdefgh”, target = “addddddd”, original = [“bcd”,“defgh”], changed = [“ddd”,“ddddd”], cost = [100,1578]
输出：-1
解释：无法将 “abcdefgh” 转换为 “addddddd” 。
如果选择子串 source[1…3] 执行第一次操作，以将 “abcdefgh” 改为 “adddefgh” ，你无法选择子串 source[3…7] 执行第二次操作，因为两次操作有一个共用下标 3 。
如果选择子串 source[3…7] 执行第一次操作，以将 “abcdefgh” 改为 “abcddddd” ，你无法选择子串 source[1…3] 执行第二次操作，因为两次操作有一个共用下标 3 。
参数范围：
1 <= source.length == target.length <= 1000
source、target 均由小写英文字母组成
1 <= cost.length == original.length == changed.length <= 100
1 <= original[i].length == changed[i].length <= source.length
original[i]、changed[i] 均由小写英文字母组成
original[i] != changed[i]
1 <= cost[i] <= 10⁶

分析

将s按顺序拆分成若干子串，任何字符都在一个子串中，且只在一个字串中。求这些子串全部转成t的最小成本。

动态规划


动态规划之状态表示	dp[i]表示将s[0,i)转化成t[0,i)的最小成本
动态规划之状态转移方程	dp[j]=min(dp[i]+s[i,j)转化成t[i,j)成本), i取值范围[0,j)
动态规划之状态之初始化	dp[0]=0
动态规划之状态之填表顺序	两层循环枚举i,j ，先枚举i,再枚举j。s[i,j)是最后一个子串
动态规划之状态之返回值	dp.back()

字符经过多次转化的最小成本

本质就是最短多源路径。

将字符串转成整数（节点编号）

如果用哈希map，每次是O(n)，就超时了。可以自写哈希或用字典树，从查询s[i,j)到s[i+j+1)时间复杂度是O(1)。总时间复杂度是O(n²)。

测试用例

template<class T>
void Assert(const T& t1, const T& t2)
{assert(t1 == t2);
}template<class T>
void Assert(const vector<T>& v1, const vector<T>& v2)
{if (v1.size() != v2.size()){assert(false);return;}for (int i = 0; i < v1.size(); i++){Assert(v1[i], v2[i]);}
}int main()
{string source, target;vector<string> original, changed;vector<int> cost;{Solution sln;source = "abcdefgh", target = "acdeeghh", original = { "bcd", "fgh", "thh" }, changed = { "cde", "thh", "ghh" }, cost = { 1, 3, 5 };auto res = sln.minimumCost(source, target, original, changed, cost);Assert(9LL, res);}{Solution sln;source = "abcd", target = "acbe";original = { "a", "b", "c", "c", "e", "d" }, changed = { "b", "c", "b", "e", "b", "e" };cost = { 2, 5, 5, 1, 2, 20 };auto res = sln.minimumCost(source, target, original, changed, cost);Assert(28LL, res);}{Solution sln;source = "abbaacddcbacbddcbdbddcadbadbbdbaabcdbdbdcaccacaddcabadadaccabbddbbdacaacdbdcaaddcacbcbcaaacaddabcabc";target = "abbaacdbcbabcadcbdbddcadbadbbddaabcdbdddcacadbacabcbdbbbbdaabbddbbdaabbcdbdcaaddcacbcbcadadccdcbcbcb";original = { "b", "c", "a", "cbd", "adc", "ddb", "dcb", "dbd", "b", "caac", "acdc", "cbbd", "bcdb", "ddbc", "aacadda", "cbadbbd", "aaabcad", "bacdccc", "ddabdaa", "abc", "bbc", "cdd", "ddb", "cacaddcabad", "bdcdccabdcb", "bddbbabdcac", "adacc", "bbdca", "dabad", "cddcba" };changed = { "c", "a", "d", "adc", "ddb", "dcb", "dbd", "bca", "d", "acdc", "cbbd", "bcdb", "ddbc", "abbc", "cbadbbd", "aaabcad", "bacdccc", "ddabdaa", "dadccdc", "bbc", "cdd", "ddb", "bcb", "bdcdccabdcb", "bddbbabdcac", "adbacabcbdb", "bbdca", "dabad", "bbbda", "cdbcba" };cost = { 63, 87, 77, 94, 100, 73, 99, 16, 89, 94, 76, 43, 76, 84, 83, 38, 96, 87, 34, 98, 33, 53, 58, 90, 61, 51, 92, 76, 91, 70 };auto res = sln.minimumCost(source, target, original, changed, cost);Assert(2044LL, res);}{Solution sln;source = "aaaddcaaccbabaaccbabbaadcccadbaacbddbaccabddbdbaaddbbacbddddbbdbccaadcaccacdbcbddbacabadaaccbadbbdbc";target = "abaddcabcdbabcbadcaccaadabbadddbcacaaabdabbdcbbdbcbaaabbbcddcbddcbccadacddcbdcbacadbbadbdabcbadbbdac";original = { "ddddb", "dccbb", "dadac", "dbdbb", "ddbacabadaac", "bcbccdcadabd", "dacabcdaacca", "dcdadacacbbd", "dcccadbaacbddbacc", "dcdcbccdccdbaaaac", "bbbcccdbcdcadaabc", "bccaadcaccacdb", "bbcabcbcbaddbd", "dbadadaddcddad", "badaddbcddacca", "bc", "da", "cb", "ddbdbaaddbbac", "dbcadcdbabddd", "abdadacbbbcca", "adaaabcabdbcc", "caaccbabaaccbabba", "abaadddbaaccbbacc", "bbddaaadcbccccbac", "cdbdbddaadbbbdbdd", "bcbdaabaddbdcdcaa", "aa", "cb", "dd" };changed = { "dccbb", "dadac", "dbdbb", "bcddc", "bcbccdcadabd", "dacabcdaacca", "dcdadacacbbd", "acadbbadbdab", "dcdcbccdccdbaaaac", "bbbcccdbcdcadaabc", "dabbadddbcacaaabd", "bbcabcbcbaddbd", "dbadadaddcddad", "badaddbcddacca", "dcbccadacddcbd", "da", "cb", "ac", "dbcadcdbabddd", "abdadacbbbcca", "adaaabcabdbcc", "bdcbbdbcbaaab", "abaadddbaaccbbacc", "bbddaaadcbccccbac", "cdbdbddaadbbbdbdd", "bcbdaabaddbdcdcaa", "cabcdbabcbadcacca", "cb", "dd", "ba" };cost = { 67, 56, 64, 83, 100, 73, 95, 97, 100, 98, 20, 92, 58, 70, 95, 77, 95, 93, 69, 92, 77, 53, 96, 68, 83, 96, 93, 64, 81, 100 };auto res = sln.minimumCost(source, target, original, changed, cost);Assert(2405LL, res);}//CConsole::Out(res);
}

代码

我写了N版都超时，单个用例不超时，总用例超时。用题解的代码运行了错误和超时用例，速度比我的速度快了近一半。算了，不继续优化了，许多比赛的技巧是工作的灾难，比如用原生数组代替vector。

第六版超时

template<class TData, TData defData,int iTypeNum = 26, TData cBegin = 'a'>
class CTrie
{
public:CTrie() {m_iID = s_ID++;}int GetLeadCount(){return m_iLeafCount;}template<class IT>int Add(IT begin, IT end){int iLeve = 0;CTrie* pNode = this;for (; begin != end; ++begin){pNode = pNode->AddChar(*begin);			pNode->m_iLeve = iLeve++;}if (-1 == pNode->m_iLeafID){pNode->m_iLeafID = m_iLeafCount++;}return pNode->m_iLeafID;}template<class IT>CTrie* Search(IT begin, IT end){if (begin == end){return this;}if ('.' == *begin){for (auto& ptr : m_vPChilds){if (!ptr){continue;}auto pSearch = ptr->Search(begin + 1, end);if (pSearch){return pSearch;}}return nullptr;}auto ptr = GetChild(*begin);if (nullptr == ptr){return nullptr;}return ptr->Search(begin + 1, end);}TData m_data = defData;CTrie* AddChar(TData ele){if ((ele < cBegin) || (ele >= cBegin + iTypeNum)){return nullptr;}const int index = ele - cBegin;auto ptr = m_vPChilds[index];if (!ptr){m_vPChilds[index] = new CTrie();}return m_vPChilds[index];}CTrie* GetChild(TData ele)const{if ((ele < cBegin) || (ele >= cBegin + iTypeNum)){return nullptr;}return m_vPChilds[ele - cBegin];}
protected:int m_iID;
public:int m_iLeafID=-1;
protected:int m_iLeve=-1;inline static int s_ID = 0;int m_iLeafCount = 0;CTrie* m_vPChilds[iTypeNum] = { nullptr };
};template<char defData='a', int iTypeNum = 26, char cBegin = 'a'>
class CStrTrieHelp 
{
public:int Add(const string& s){return m_trie.Add(s.begin(), s.begin() + s.length());}CTrie<char, defData, iTypeNum, cBegin>* Search(const string& s){return m_trie.Search(s.begin(), s.begin() + s.length());}CTrie<char, defData, iTypeNum, cBegin>* SearchSub(const string& s,int iPos,int len){return m_trie.Search(s.begin()+ iPos, s.begin() + iPos + len );}CTrie<char, defData, iTypeNum, cBegin> m_trie;
};
template<char defData = 'a', int iTypeNum = 26, char cBegin = 'a'>
class CStrIndexs
{
public:void Add(const string& s){m_trie.Add(s);}int Seach(const string& s){auto p = m_trie.Search(s);if (nullptr == p){return -1;}return p->m_iLeafID;}int SearchSub(const string& s, int iPos, int len){auto p = m_trie.SearchSub(s, iPos, len);if (nullptr == p){return -1;}return p->m_iDebug;}CStrTrieHelp<defData, iTypeNum, cBegin> m_trie;
};//多源码路径
template<class T,T INF = 1000*1000*1000>
class CFloyd
{
public:CFloyd(const  vector<vector<T>>& mat){m_vMat = mat;const int n = mat.size();for (int i = 0; i < n; i++){//通过i中转for (int i1 = 0; i1 < n; i1++){for (int i2 = 0; i2 < n; i2++){//此时：m_vMat[i1][i2] 表示通过[0,i)中转的最短距离m_vMat[i1][i2] = min(m_vMat[i1][i2], m_vMat[i1][i] + m_vMat[i][i2]);//m_vMat[i1][i2] 表示通过[0,i]中转的最短距离}}}		};vector<vector<T>> m_vMat;
};class Solution {
public:long long minimumCost(string source, string target, vector<string>& original, vector<string>& changed, vector<int>& cost) {CStrIndexs strIndexs;for (const auto& s : original){strIndexs.Add(s);}for (const auto& s : changed){strIndexs.Add(s);}const int iNotMay = 1000 * 1000 * 1000;vector<vector<int>> vMat(strIndexs.m_trie.m_trie.GetLeadCount(), vector<int>(strIndexs.m_trie.m_trie.GetLeadCount(), iNotMay));for (int j = 0; j < original.size(); j++){const int iSrc = strIndexs.Seach(original[j]);const int iDest = strIndexs.Seach(changed[j]);			auto& n = vMat[iSrc][iDest];n = min(n, cost[j]);}for (int i = 0; i < strIndexs.m_trie.m_trie.GetLeadCount(); i++){vMat[i][i] = 0;}CFloyd floyd(vMat);vector<long long> vRet(source.length() + 1, LLONG_MAX / 1000);vRet[0] = 0;for (int i = 0; i < source.length(); i++){bool bSame = true;auto pSrc = &strIndexs.m_trie.m_trie;auto pDst = &strIndexs.m_trie.m_trie;for (int len = 1; len + i <= source.length(); len++){const char ch1 = source[len + i - 1];const char ch2 = target[len + i - 1];				bSame &= (ch1 == ch2);if (nullptr != pSrc){pSrc = pSrc->GetChild(ch1);}if (nullptr != pDst){pDst = pDst->GetChild(ch2);}if (bSame){vRet[i + len] = min(vRet[i + len], vRet[i]);continue;}					if ((nullptr == pSrc) || (nullptr == pDst)){break;}const int iSrc = pSrc->m_iLeafID;const int iDest = pDst->m_iLeafID;if ((-1 == iSrc) || (-1== iDest)){continue;}const int iDist = floyd.m_vMat[iSrc][iDest];if (iDist >= iNotMay){continue;}vRet[i + len] = min(vRet[i + len], vRet[i] + iDist);}}return (vRet.back() >= LLONG_MAX / 1000) ? -1 : vRet.back();}
};

第一版超时

//多源码路径
template<class T,T INF = 100010001000>
class CFloyd
{
public:
CFloyd(const vector<vector>& mat)
{
m_vMat = mat;
const int n = mat.size();
for (int i = 0; i < n; i++)
{//通过i中转
for (int i1 = 0; i1 < n; i1++)
{
for (int i2 = 0; i2 < n; i2++)
{
//此时：m_vMat[i1][i2] 表示通过[0,i)中转的最短距离
m_vMat[i1][i2] = min(m_vMat[i1][i2], m_vMat[i1][i] + m_vMat[i][i2]);
//m_vMat[i1][i2] 表示通过[0,i]中转的最短距离
}
}
}
};
vector<vector> m_vMat;
};

class Solution {
public:
long long minimumCost(string source, string target, vector& original, vector& changed, vector& cost) {
vector strs(original.begin(), original.end());
strs.insert(strs.end(), changed.begin(), changed.end());
sort(strs.begin(),strs.end());
strs.erase(std::unique(strs.begin(), strs.end()), strs.end());
std::unordered_map<string, int> mStrToNode;
for (int i = 0; i < strs.size(); i++)
{
mStrToNode[strs[i]] = i;
}
const int iNotMay = 1000 * 1000 * 1000;
vector<vector> vMat(strs.size(), vector(strs.size(), iNotMay));
vector vOriNode;
for (int j = 0; j < original.size(); j++)
{
vOriNode.emplace_back(mStrToNode[original[j]]);
auto& n = vMat[vOriNode.back()][mStrToNode[changed[j]]];
n = min(n,cost[j]);
}
for (int i = 0; i < strs.size(); i++)
{
vMat[i][i] = 0;
}
CFloyd floyd(vMat);
vector vRet(source.length() + 1,LLONG_MAX/1000 );
vRet[0]=0;
for (int i = 0; i < source.length(); i++)
{
if (source[i] == target[i])
{
vRet[i + 1] = min(vRet[i+1],vRet[i]);
//continue; 相等也可以替换
}
for (int j = 0; j < original.size(); j++)
{
const int len = original[j].length();
if (i + len > source.length())
{
continue;
}
if (source.substr(i, len) != original[j])
{
continue;
}
string sDst = target.substr(i, len);
if (!mStrToNode.count(sDst))
{
continue;
}
const int iDest = mStrToNode[sDst];
const int iDist = floyd.m_vMat[vOriNode[j]][iDest];
if (iDist >= iNotMay)
{
continue;
}
vRet[i + len] = min(vRet[i + len],vRet[i]+iDist);
}
}
return (vRet.back() >= LLONG_MAX / 1000) ? -1 : vRet.back();
}
};

第二版超时

class Solution {
public:
long long minimumCost(string source, string target, vector& original, vector& changed, vector& cost) {
vector strs(original.begin(), original.end());
strs.insert(strs.end(), changed.begin(), changed.end());
sort(strs.begin(),strs.end());
strs.erase(std::unique(strs.begin(), strs.end()), strs.end());
std::unordered_map<string, int> mStrToNode;
for (int i = 0; i < strs.size(); i++)
{
mStrToNode[strs[i]] = i;
}
const int iNotMay = 1000 * 1000 * 1000;
vector<vector> vMat(strs.size(), vector(strs.size(), iNotMay));
for (int j = 0; j < original.size(); j++)
{
auto& n = vMat[mStrToNode[original[j]]][mStrToNode[changed[j]]];
n = min(n,cost[j]);
}
for (int i = 0; i < strs.size(); i++)
{
vMat[i][i] = 0;
}
CFloyd floyd(vMat);
vector vRet(source.length() + 1,LLONG_MAX/1000 );
vRet[0]=0;
for (int i = 0; i < source.length(); i++)
{
for (int len = 1; len + i <= source.length(); len++)
{
const string sSrc = source.substr(i, len);
const string sDst = target.substr(i, len);
if (sSrc == sDst)
{
vRet[i + len] = min(vRet[i + len], vRet[i] );
continue;
}
if (!mStrToNode.count(sDst)|| !mStrToNode.count(sSrc))
{
continue;
}
const int iSrc = mStrToNode[sSrc];
const int iDest = mStrToNode[sDst];
const int iDist = floyd.m_vMat[iSrc][iDest];
if (iDist >= iNotMay)
{
continue;
}
vRet[i + len] = min(vRet[i + len], vRet[i] + iDist);
}
}
return (vRet.back() >= LLONG_MAX / 1000) ? -1 : vRet.back();
}
};

第四版超时

template<class TData, TData defData,int iTypeNum = 26, TData cBegin = ‘a’>
class CTrie
{
public:
CTrie()
{

}
template<class IT>
CTrie* Add(IT begin, IT end,const int iDebug)
{int iLeve = 0;CTrie* pNode = this;for (; begin != end; ++begin){pNode = pNode->AddChar(*begin);			pNode->m_iLeve = iLeve++;}pNode->m_iDebug = iDebug;return pNode;
}
template<class IT>
CTrie* Search(IT begin, IT end)
{if (begin == end){return this;}if ('.' == *begin){for (auto& ptr : m_vPChilds){if (!ptr){continue;}auto pSearch = ptr->Search(begin + 1, end);if (pSearch){return pSearch;}}return nullptr;}auto ptr = GetChild(*begin);if (nullptr == ptr){return nullptr;}return ptr->Search(begin + 1, end);
}
TData m_data = defData;CTrie* AddChar(TData ele)
{if ((ele < cBegin) || (ele >= cBegin + iTypeNum)){return nullptr;}const int index = ele - cBegin;auto ptr = m_vPChilds[index];if (!ptr){m_vPChilds[index] = new CTrie();}return m_vPChilds[index];
}CTrie* GetChild(TData ele)const
{if ((ele < cBegin) || (ele >= cBegin + iTypeNum)){return nullptr;}return m_vPChilds[ele - cBegin];
}int m_iDebug=-1;

protected:
int m_iLeve=-1;
CTrie* m_vPChilds[iTypeNum] = { nullptr };
};

template<char defData, int iTypeNum = 26, char cBegin = ‘a’>
class CStrTrieHelp
{
public:
CTrie<char, defData, iTypeNum, cBegin>* Add(const string& s,int iDebug)
{
return m_trie.Add(s.begin(), s.begin() + s.length(), iDebug);
}
CTrie<char, defData, iTypeNum, cBegin>* Search(const string& s)
{
return m_trie.Search(s.begin(), s.begin() + s.length());
}
CTrie<char, defData, iTypeNum, cBegin>* SearchSub(const string& s,int iPos,int len)
{
return m_trie.Search(s.begin()+ iPos, s.begin() + iPos + len );
}
CTrie<char, defData, iTypeNum, cBegin> m_trie;
};
template<char defData = ‘a’, int iTypeNum = 26, char cBegin = ‘a’>
class CStrIndexs
{
public:
void Add(const string& s, int iDebug)
{
m_trie.Add(s, iDebug);
}
int Seach(const string& s)
{
auto p = m_trie.Search(s);
if (nullptr == p)
{
return -1;
}
return p->m_iDebug;
}
int SearchSub(const string& s, int iPos, int len)
{
auto p = m_trie.SearchSub(s, iPos, len);
if (nullptr == p)
{
return -1;
}
return p->m_iDebug;
}
protected:
CStrTrieHelp<defData, iTypeNum, cBegin> m_trie;
};

class Solution {
public:
long long minimumCost(string source, string target, vector& original, vector& changed, vector& cost) {
vector strs(original.begin(), original.end());
strs.insert(strs.end(), changed.begin(), changed.end());
sort(strs.begin(), strs.end());
strs.erase(std::unique(strs.begin(), strs.end()), strs.end());
CStrIndexs strIndexs;
for (int i = 0; i < strs.size(); i++)
{
strIndexs.Add(strs[i], i);
}
const int iNotMay = 1000 * 1000 * 1000;
vector<vector> vMat(strs.size(), vector(strs.size(), iNotMay));
for (int j = 0; j < original.size(); j++)
{
const int iSrc = strIndexs.Seach(original[j]);
const int iDest = strIndexs.Seach(changed[j]);
auto& n = vMat[iSrc][iDest];
n = min(n, cost[j]);
}
for (int i = 0; i < strs.size(); i++)
{
vMat[i][i] = 0;
}
CFloyd floyd(vMat);
vector vRet(source.length() + 1, LLONG_MAX / 1000);
vRet[0] = 0;
for (int i = 0; i < source.length(); i++)
{
bool bSame = true;
for (int len = 1; len + i <= source.length(); len++)
{
bSame &= (source[len + i - 1] == target[len + i - 1]);
if (bSame)
{
vRet[i + len] = min(vRet[i + len], vRet[i]);
continue;
}
const int iSrc = strIndexs.SearchSub(source, i, len);
const int iDest = strIndexs.SearchSub(target, i, len);
if ((-1 == iSrc) || (-1== iDest))
{
continue;
}
const int iDist = floyd.m_vMat[iSrc][iDest];
if (iDist >= iNotMay)
{
continue;
}
vRet[i + len] = min(vRet[i + len], vRet[i] + iDist);
}
}
return (vRet.back() >= LLONG_MAX / 1000) ? -1 : vRet.back();
}
};

扩展阅读

视频课程

有效学习：明确的目标及时的反馈拉伸区（难度合适），可以先学简单的课程，请移步CSDN学院，听白银讲师（也就是鄙人）的讲解。
https://edu.csdn.net/course/detail/38771

如何你想快

速形成战斗了，为老板分忧，请学习C#入职培训、C++入职培训等课程
https://edu.csdn.net/lecturer/6176

我想对大家说的话
闻缺陷则喜是一个美好的愿望，早发现问题，早修改问题，给老板节约钱。
子墨子言之：事无终始，无务多业。也就是我们常说的专业的人做专业的事。
如果程序是一条龙，那算法就是他的是睛

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。