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
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 }