牛客找到二叉树中符合搜索二叉树条件的最大拓扑结构

小灰灰 2023-06-04 14:53 3阅读 0赞

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][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 ###

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1 #include <bits/stdc++.h>
      2 using namespace std;
      3  
      4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
      5 #define Rep(i,n) for (int i = 0; i < (int)(n); ++i)
      6 #define For(i,s,t) for (int i = (int)(s); i <= (int)(t); ++i)
      7 #define rFor(i,t,s) for (int i = (int)(t); i >= (int)(s); --i)
      8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
      9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
     10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
     11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
     12  
     13 #define pr(x) cout << #x << " = " << x << "  "
     14 #define prln(x) cout << #x << " = " << x << endl
     15  
     16 #define LOWBIT(x) ((x)&(-x))
     17  
     18 #define ALL(x) x.begin(),x.end()
     19 #define INS(x) inserter(x,x.begin())
     20 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
     21 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 
     22 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
     23 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);
     24  
     25 #define ms0(a) memset(a,0,sizeof(a))
     26 #define msI(a) memset(a,0x3f,sizeof(a))
     27 #define msM(a) memset(a,-1,sizeof(a))
     28 
     29 #define MP make_pair
     30 #define PB push_back
     31 #define ft first
     32 #define sd second
     33  
     34 template<typename T1, typename T2>
     35 istream &operator>>(istream &in, pair<T1, T2> &p) {
     36     in >> p.first >> p.second;
     37     return in;
     38 }
     39  
     40 template<typename T>
     41 istream &operator>>(istream &in, vector<T> &v) {
     42     for (auto &x: v)
     43         in >> x;
     44     return in;
     45 }
     46 
     47 template<typename T>
     48 ostream &operator<<(ostream &out, vector<T> &v) {
     49     Rep(i, v.size()) out << v[i] << " \n"[i == v.size() - 1];
     50     return out;
     51 }
     52  
     53 template<typename T1, typename T2>
     54 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     55     out << "[" << p.first << ", " << p.second << "]" << "\n";
     56     return out;
     57 }
     58 
     59 inline int gc(){
     60     static const int BUF = 1e7;
     61     static char buf[BUF], *bg = buf + BUF, *ed = bg;
     62     
     63     if(bg == ed) fread(bg = buf, 1, BUF, stdin);
     64     return *bg++;
     65 } 
     66 
     67 inline int ri(){
     68     int x = 0, f = 1, c = gc();
     69     for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
     70     for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
     71     return x*f;
     72 }
     73 
     74 template<class T>
     75 inline string toString(T x) {
     76     ostringstream sout;
     77     sout << x;
     78     return sout.str();
     79 }
     80 
     81 inline int toInt(string s) {
     82     int v;
     83     istringstream sin(s);
     84     sin >> v;
     85     return v;
     86 }
     87 
     88 //min <= aim <= max
     89 template<typename T>
     90 inline bool BETWEEN(const T aim, const T min, const T max) {
     91     return min <= aim && aim <= max;
     92 }
     93  
     94 typedef long long LL;
     95 typedef unsigned long long uLL;
     96 typedef vector< int > VI;
     97 typedef vector< bool > VB;
     98 typedef vector< char > VC;
     99 typedef vector< double > VD;
    100 typedef vector< string > VS;
    101 typedef vector< LL > VL;
    102 typedef vector< VI > VVI;
    103 typedef vector< VB > VVB;
    104 typedef vector< VS > VVS;
    105 typedef vector< VL > VVL;
    106 typedef vector< VVI > VVVI;
    107 typedef vector< VVL > VVVL;
    108 typedef pair< int, int > PII;
    109 typedef pair< LL, LL > PLL;
    110 typedef pair< int, string > PIS;
    111 typedef pair< string, int > PSI;
    112 typedef pair< string, string > PSS;
    113 typedef pair< double, double > PDD;
    114 typedef vector< PII > VPII;
    115 typedef vector< PLL > VPLL;
    116 typedef vector< VPII > VVPII;
    117 typedef vector< VPLL > VVPLL;
    118 typedef vector< VS > VVS;
    119 typedef map< int, int > MII;
    120 typedef unordered_map< int, int > uMII;
    121 typedef map< LL, LL > MLL;
    122 typedef map< string, int > MSI;
    123 typedef map< int, string > MIS;
    124 typedef set< int > SI;
    125 typedef stack< int > SKI;
    126 typedef deque< int > DQI;
    127 typedef queue< int > QI;
    128 typedef priority_queue< int > PQIMax;
    129 typedef priority_queue< int, VI, greater< int > > PQIMin;
    130 const double EPS = 1e-8;
    131 const LL inf = 0x7fffffff;
    132 const LL infLL = 0x7fffffffffffffffLL;
    133 const LL mod = 1e9 + 7;
    134 const int maxN = 2e5 + 7;
    135 const LL ONE = 1;
    136 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
    137 const LL oddBits = 0x5555555555555555;
    138 
    139 struct TreeNode {
    140     int lch = 0, rch = 0, val = 0;
    141 };
    142 
    143 int N, root, ans;
    144 TreeNode tree[maxN];
    145 
    146 // 查找以rt为根的二叉树中最右的第一个比k大的节点 
    147 // 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 
    148 int getMax(int rt, int k) {
    149     if(!rt) return 0;
    150     while(tree[rt].rch && tree[rt].rch < k && rt < tree[rt].rch) rt = tree[rt].rch;
    151     if(rt < tree[rt].rch) return tree[rt].rch;
    152     return 0;
    153 }
    154 
    155 // 查找以rt为根的二叉树中最左的第一个比k小的节点 
    156 int getMin(int rt, int k) {
    157     if(!rt) return 0;
    158     while(tree[rt].lch && tree[rt].lch > k && tree[rt].lch < rt) rt = tree[rt].lch;
    159     if(tree[rt].lch < rt) return tree[rt].lch;
    160     return 0;
    161 }
    162 
    163 void dfs(int rt) {
    164     if(!rt) return;
    165     tree[rt].val = 1;
    166     
    167     dfs(tree[rt].lch);
    168     dfs(tree[rt].rch);
    169     
    170     int mostRight = getMax(tree[rt].lch, rt);
    171     int mostLeft = getMin(tree[rt].rch, rt);
    172     
    173     if(tree[rt].lch && tree[rt].lch < rt) {
    174         tree[rt].val += tree[tree[rt].lch].val;
    175         if(mostRight) {
    176             tree[rt].val -= tree[mostRight].val;
    177             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 
    178         }
    179     }
    180     if(tree[rt].rch && tree[rt].rch > rt) {
    181         tree[rt].val += tree[tree[rt].rch].val;
    182         if(mostLeft) {
    183             tree[rt].val -= tree[mostLeft].val;
    184             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  
    185         }
    186     }
    187     
    188     ans = max(ans, tree[rt].val);
    189 }
    190 
    191 int main(){
    192     //freopen("MyOutput.txt","w",stdout);
    193     //freopen("input.txt","r",stdin);
    194     //INIT();
    195     scanf("%d%d", &N, &root);
    196     Rep(i, N) {
    197         int fa, lch, rch;
    198         scanf("%d%d%d", &fa, &lch, &rch);
    199         
    200         tree[fa].lch = lch;
    201         tree[fa].rch = rch;
    202     }
    203     
    204     dfs(root);
    205     
    206     printf("%d\n", ans);
    207     return 0;
    208 }

转载于:https://www.cnblogs.com/zaq19970105/p/11359966.html

[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]: 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
[ContractedBlock.gif]: https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif
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