Cattle off to find two nodes in the binary tree in a recent common ancestor (Advanced)

Topic links: https://www.nowcoder.com/practice/357a9856c629405a8405d293bd6be2d1?tpId=101&tqId=33245&tPage=1&rp=1&ru=/ta/programmer-code-interview-guide&qru=/ta/programmer-code-interview-guide/question -ranking

Subject to the effect

  slightly.

analysis

code show as below

  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<classT>
 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 in;
83      istringstream son (s);
84      son >> in;
85      return to;
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 = 1e3 + 7;
135 const LL ONE = . 1 ;
 136  const LL evenBits = 0xaaaaaaaaaaaaaaaa ;
 137  const LL oddBits = 0x5555555555555555 ;
 138  
139  struct the TreeNode {
 140      int LCH = 0 , Rch = 0 , Val = 0 , FA = 0 , Level = 0 ;
 141 is  };
 142  
143  int N, M, the root, O1, O2;
 144  the TreeNode tree [maxN];
 145  
146 Vl Euler; // tree node Euler sequence 
147Depth Vl; // Euler sequence corresponding to the depth of node 
148  int firstpos [maxN]; // the position of each node appears first in the sequence of Euler 
 149  
150  // ST [I] [J] indicates an array interval Euler sequence [i, i + 2 ^ j - 1] corresponding to the minimum value of the Euler sequence index 
151  int ST [maxN * . 3 ] [ 10 ];
 152  
153  // Euler sequence 
154 inline void DFS ( int RT, int D) {
 155      firstpos [RT] = ( int ) euler.size ();
 156      euler.PB (RT);
 157      depth.PB (D);
 158  
159      IF(tree[rt].lch) {
160         dfs(tree[rt].lch, d + 1);
161         euler.PB(rt);
162         depth.PB(d);
163     }
164     if(tree[rt].rch) {
165         dfs(tree[rt].rch, d + 1);
166         euler.PB(rt);
167         depth.PB(d);
168     }
169 }
170 
171 void ST(int n) {
172     For(i, 1, n) st[i][0] = i;
173     
174     for(int j = 1; (1 << j) <= n; ++j) {
175         for(int i = 1; i + (1 << j) - 1 <= n; ++i) {
176             int a = st[i][j - 1];
177             int b = st[i + (1 << (j - 1))][j - 1];
178             st[i][j] = depth[a] <= depth[b] ? a : b;
179         }
180     }
181 }
182 
183 // [l, 2^k - 1] 和 [r - 2^k + 1, r] 可能有交叉 
184 inline int RMQ(int l, int r) {
185     int k = 32 - __builtin_clz((unsigned int)(r - l + 1)) - 1;
186     
187     int a = st[l][k];
188     int b = st[r - (1 << k) + 1][k];
189     return depth[a] <= depth[b] ? a : b;
190 }
191  
192 int LCA(int o1, int o2) {
193     int x = firstPos[o1] , y = firstPos[o2];
194     if(x > y) swap(x, y);
195     return euler[RMQ(x, y)];
196 }
197 
198 
199 int main(){
200     //freopen("MyOutput.txt","w",stdout);
201     //freopen("input.txt","r",stdin);
202     //INIT();
203     scanf("%d%d", &N, &root);
204     Rep(i, N) {
205         int fa, lch, rch;
206         scanf("%d%d%d", &fa, &lch, &rch);
207         
208         tree[fa].lch = lch;
209         tree[fa].rch = rch;
210     }
211     
212     euler.PB(-1);
213     depth.PB(-1);
214     dfs(root, 1);
215     ST((int)euler.size() - 1);
216     
217     scanf("%d", &M);
218     Rep(i, M) {
219         scanf("%d%d", &o1, &o2);
220         printf("%d\n", LCA(o1, o2));
221     }
222     return 0;
223 }
View Code

 

Guess you like

Origin www.cnblogs.com/zaq19970105/p/11367027.html