SPOJ 16549 - QTREE6 - Query on a tree VI 「一种维护树上颜色连通块的操作」

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 
  5 using namespace std;
  6 
  7 const int MAXN = 1e06 + 10;
  8 const int MAXM = 1e06 + 10;
  9 
 10 struct LinkedForwardStar {
 11     int to;
 12 
 13     int next;
 14 } ;
 15 
 16 LinkedForwardStar Link[MAXM << 1];
 17 int Head[MAXN]= {0};
 18 int size = 0;
 19 
 20 void Insert (int u, int v) {
 21     Link[++ size].to = v;
 22     Link[size].next = Head[u];
 23 
 24     Head[u] = size;
 25 }
 26 
 27 int N, M;
 28 int ances[MAXN];
 29 
 30 struct Link_Cut_Tree {
 31     int father[MAXN];
 32     int son[MAXN][2];
 33     int subtree[MAXN], virsize[MAXN];
 34 
 35     void init () {
 36         for (int i = 1; i <= N; i ++)
 37             father[i] = son[i][0] = son[i][1] = subtree[i] = virsize[i] = 0;
 38     }
 39     int isroot (int p) {
 40         return son[father[p]][0] != p && son[father[p]][1] != p;
 41     }
 42     int sonbel (int p) {
 43         return son[father[p]][1] == p;
 44     }
 45     void pushup (int p) {
 46         subtree[p] = subtree[son[p][0]] + subtree[son[p][1]] + virsize[p] + 1;
 47     }
 48     void rotate (int p) {
 49         int fa = father[p], anc = father[fa];
 50         int s = sonbel (p);
 51         son[fa][s] = son[p][s ^ 1];
 52         if (son[fa][s])
 53             father[son[fa][s]] = fa;
 54         if (! isroot (fa))
 55             son[anc][sonbel (fa)] = p;
 56         father[p] = anc;
 57         son[p][s ^ 1] = fa, father[fa] = p;
 58         pushup (fa), pushup (p);
 59     }
 60     void splay (int p) {
 61         for (int fa = father[p]; ! isroot (p); rotate (p), fa = father[p])
 62             if (! isroot (fa))
 63                 sonbel (p) == sonbel (fa) ? rotate (fa) : rotate (p);
 64     }
 65     void Access (int p) {
 66         for (int tp = 0; p; tp = p, p = father[p]) {
 67             splay (p);
 68             virsize[p] += subtree[son[p][1]];
 69             son[p][1] = tp;
 70             virsize[p] -= subtree[son[p][1]];
 71             pushup (p);
 72         }
 73     }
 74     int findroot (int p) {
 75         Access (p), splay (p);
 76         while (son[p][0])
 77             p = son[p][0];
 78         splay (p);
 79         return p;
 80     }
 81     void link (int p) {
 82         int fa = ances[p];
 83         splay (p);
 84         father[p] = fa;
 85         Access (fa), splay (fa);
 86         subtree[fa] += subtree[p], virsize[fa] += subtree[p];
 87     }
 88     void cut (int p) {
 89         Access (p), splay (p);
 90         father[son[p][0]] = 0, son[p][0] = 0;
 91         pushup (p);
 92     }
 93 } ;
 94 Link_Cut_Tree LCT[2];
 95 
 96 void DFS (int root, int father) {
 97     ances[root] = father;
 98     LCT[0].link (root);
 99     for (int i = Head[root]; i; i = Link[i].next) {
100         int v = Link[i].to;
101         if (v == father)
102             continue;
103         DFS (v, root);
104     }
105 }
106 
107 int Colour[MAXN]= {0};
108 
109 int getnum () {
110     int num = 0;
111     char ch = getchar ();
112 
113     while (! isdigit (ch))
114         ch = getchar ();
115     while (isdigit (ch))
116         num = (num << 3) + (num << 1) + ch - ‘0‘, ch = getchar ();
117 
118     return num;
119 }
120 
121 int main () {
122     N = getnum ();
123     for (int i = 1; i <= N; i ++)
124         LCT[0].subtree[i] = LCT[1].subtree[i] = 1;
125     for (int i = 1; i < N; i ++) {
126         int u = getnum (), v = getnum ();
127         Insert (u, v), Insert (v, u);
128     }
129     DFS (1, N + 1);
130     M = getnum ();
131     for (int Case = 1; Case <= M; Case ++) {
132         int opt = getnum (), p = getnum ();
133         int col = Colour[p];
134         if (opt == 0) {
135             int anc = LCT[col].findroot (p);
136             printf ("%d\n", LCT[col].subtree[LCT[col].son[anc][1]]);
137             // 注意，因为有可能存在两个不连通的连通快在LCT上连通，又在Access后右节点仅包含当前链
138             // 故需输出右子树信息而并非减一，否则有可能会算上另一个连通块的答案
139         }
140         else if (opt == 1)
141             LCT[col].cut (p), LCT[Colour[p] ^= 1].link (p);
142     }
143 
144     return 0;
145 }
146 
147 /*
148 5
149 1 2
150 1 3
151 1 4
152 1 5
153 3
154 0 1
155 1 1
156 0 1
157 */
158 
159 /*
160 5
161 1 2
162 2 3
163 3 4
164 4 5
165 3
166 1 1
167 1 3
168 0 1
169 */