牛客 找到二叉树中符合搜索二叉树条件的最大拓扑结构

题目链接：https://www.nowcoder.com/practice/e13bceaca5b14860b83cbcc4912c5d4a?tpId=101&tqId=33235&tPage=1&rp=1&ru=/ta/programmer-code-interview-guide&qru=/ta/programmer-code-interview-guide/question-ranking

题目大意

　　略。

分析

　　计算以每个节点为根节点的拓扑节点数最大值，Merge 时需要扣除左右孩子拓扑结构中不符合新拓扑搜索二叉树的节点数目，合并完后需要更新左右孩子在新拓扑结构中自拓扑结构的节点数目，理由是它们以后还可能会被访问到。（如果严谨一点是要更新一条边的，不过只有左右孩子节点用得到而已）

代码如下

#include <bits/stdc++.h>
using namespace std;
 
#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (int)(n); ++i)
#define For(i,s,t) for (int i = (int)(s); i <= (int)(t); ++i)
#define rFor(i,t,s) for (int i = (int)(t); i >= (int)(s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
#define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 
#define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
#define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,0x3f,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}

template<typename T>
ostream &operator<<(ostream &out, vector<T> &v) {
    Rep(i, v.size()) out << v[i] << " 
"[i == v.size() - 1];
    return out;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}

template<class T>
inline string toString(T x) {
    ostringstream sout;
    sout << x;
    return sout.str();
}

inline int toInt(string s) {
    int v;
    istringstream sin(s);
    sin >> v;
    return v;
}

//min <= aim <= max
template<typename T>
inline bool BETWEEN(const T aim, const T min, const T max) {
    return min <= aim && aim <= max;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef vector< int > VI;
typedef vector< bool > VB;
typedef vector< char > VC;
typedef vector< double > VD;
typedef vector< string > VS;
typedef vector< LL > VL;
typedef vector< VI > VVI;
typedef vector< VB > VVB;
typedef vector< VS > VVS;
typedef vector< VL > VVL;
typedef vector< VVI > VVVI;
typedef vector< VVL > VVVL;
typedef pair< int, int > PII;
typedef pair< LL, LL > PLL;
typedef pair< int, string > PIS;
typedef pair< string, int > PSI;
typedef pair< string, string > PSS;
typedef pair< double, double > PDD;
typedef vector< PII > VPII;
typedef vector< PLL > VPLL;
typedef vector< VPII > VVPII;
typedef vector< VPLL > VVPLL;
typedef vector< VS > VVS;
typedef map< int, int > MII;
typedef unordered_map< int, int > uMII;
typedef map< LL, LL > MLL;
typedef map< string, int > MSI;
typedef map< int, string > MIS;
typedef set< int > SI;
typedef stack< int > SKI;
typedef deque< int > DQI;
typedef queue< int > QI;
typedef priority_queue< int > PQIMax;
typedef priority_queue< int, VI, greater< int > > PQIMin;
const double EPS = 1e-8;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 2e5 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

struct TreeNode {
    int lch = 0, rch = 0, val = 0;
};

int N, root, ans;
TreeNode tree[maxN];

// 查找以rt为根的二叉树中最右的第一个比k大的节点 
// 如果一直遍历到一个节点不属于以rt为根的拓扑结构，就返回 0，说明以rt为根的拓扑结构中所有节点都能融入其父节点的拓扑结构 
int getMax(int rt, int k) {
    if(!rt) return 0;
    while(tree[rt].rch && tree[rt].rch < k && rt < tree[rt].rch) rt = tree[rt].rch;
    if(rt < tree[rt].rch) return tree[rt].rch;
    return 0;
}

// 查找以rt为根的二叉树中最左的第一个比k小的节点 
int getMin(int rt, int k) {
    if(!rt) return 0;
    while(tree[rt].lch && tree[rt].lch > k && tree[rt].lch < rt) rt = tree[rt].lch;
    if(tree[rt].lch < rt) return tree[rt].lch;
    return 0;
}

void dfs(int rt) {
    if(!rt) return;
    tree[rt].val = 1;
    
    dfs(tree[rt].lch);
    dfs(tree[rt].rch);
    
    int mostRight = getMax(tree[rt].lch, rt);
    int mostLeft = getMin(tree[rt].rch, rt);
    
    if(tree[rt].lch && tree[rt].lch < rt) {
        tree[rt].val += tree[tree[rt].lch].val;
        if(mostRight) {
            tree[rt].val -= tree[mostRight].val;
            tree[tree[rt].lch].val -= tree[mostRight].val; // 左孩子以后可能还会访问到，所以要更新，其他节点不会访问到了 
        }
    }
    if(tree[rt].rch && tree[rt].rch > rt) {
        tree[rt].val += tree[tree[rt].rch].val;
        if(mostLeft) {
            tree[rt].val -= tree[mostLeft].val;
            tree[tree[rt].rch].val -= tree[mostLeft].val; // 右孩子以后可能还会访问到，所以要更新，其他节点不会访问到了  
        }
    }
    
    ans = max(ans, tree[rt].val);
}

int main(){
    //freopen("MyOutput.txt","w",stdout);
    //freopen("input.txt","r",stdin);
    //INIT();
    scanf("%d%d", &N, &root);
    Rep(i, N) {
        int fa, lch, rch;
        scanf("%d%d%d", &fa, &lch, &rch);
        
        tree[fa].lch = lch;
        tree[fa].rch = rch;
    }
    
    dfs(root);
    
    printf("%d
", ans);
    return 0;
}