# 【BZOJ2049】【LCT】Cave 洞穴勘测

Description

Input

Output

Sample Input

样例输入1 cave.in200 5Query 123 127Connect 123 127Query 123 127Destroy 127 123Query 123 127样例输入2 cave.in3 5Connect 1 2Connect 3 1Query 2 3Destroy 1 3Query 2 3

Sample Output

样例输出1 cave.outNoYesNo样例输出2 cave.outYesNo

Hint

【分析】

  1 /*
2 唐代白居易
3 《白云泉》
4 天平山上白云泉，云自无心水自闲。
5 何必奔冲山下去，更添波浪向人间。
6 */
7 #include <iostream>
8 #include <cstdio>
9 #include <algorithm>
10 #include <cstring>
11 #include <vector>
12 #include <utility>
13 #include <iomanip>
14 #include <string>
15 #include <cmath>
16 #include <queue>
17 #include <assert.h>
18 #include <map>
19 #include <ctime>
20 #include <cstdlib>
21 #include <stack>
22 #define LOCAL
23 const int INF = 0x7fffffff;
24 const int MAXN = 10000  + 10;
25 const int maxnode = 1000000;
26 const int maxm= 30000 * 2 + 10;
27 using namespace std;
28
30        struct Node{
31               int delta;//翻转标记
32               int val;//用来debug
33               Node *parent;
34               Node *ch[2];
35        }node[MAXN], *cur, _nill, *nill;
36        Node *tmp[MAXN];
37
38        bool is_root(Node* t){//判断是否是splay的根
39             if (t == nill || (t->parent->ch[0] != t && t->parent->ch[1] != t)) return 1;
40             return 0;
41        }
42        //这种题还要打标记...
43        void pushdown(Node *t){
44             if (t == nill) return ;
45             if (t->delta){
46                t->delta = 0;
47                if (t->ch[0] != nill) t->ch[0]->delta ^= 1;
48                if (t->ch[1] != nill) t->ch[1]->delta ^= 1;
49                swap(t->ch[0], t->ch[1]);
50             }
51             return;
52        }
53        void init(){
54             nill = &_nill;
55             _nill.val = 0;
56             _nill.parent = _nill.ch[0] = _nill.ch[1] = nill;
57
58             cur = node + 1;
59        }
60        Node* NEW(int val){
61             cur->parent = cur->ch[0] = cur->ch[1] = nill;
62             cur->delta = 0;
63             cur->val = val;
64             return cur++;
65        }
66        void rotate(Node *t, int d){
67             if (is_root(t)) return;
68             Node *p = t->parent;
69             p->ch[d ^ 1] = t->ch[d];
70             if (t->ch[d] != nill) t->ch[d]->parent = p;
71             t->parent = p->parent;
72             if (p != nill){
73                if (p->parent->ch[0] == p) p->parent->ch[0] = t;
74                else if (p->parent->ch[1] == p) p->parent->ch[1] = t;
75             }
76             t->ch[d] = p;
77             p->parent = t;
78             //update(p);真逗，什么都不要
79        }
80        //将t旋转到根
81        void splay(Node *t){
82             //标记下传
83             int cnt = 1;
84             tmp[0] = t;
85             for (Node *y = t; !is_root(y); y = y->parent) tmp[cnt++] = y->parent;
86             while (cnt) pushdown(tmp[--cnt]);
87
88             while (!is_root(t)){
89                   Node *y = t->parent;
90                   if (is_root(y)) rotate(t, (y->ch[0] == t));
91                   else {
92                        int d = (y->parent->ch[0] == y);
93                        if (y->ch[d] == t) rotate(t, d ^ 1);
94                        else rotate(y, d);
95                        rotate(t, d);
96                   }
97             }
98        }
99        Node* access(Node *t){
100             Node *p = nill;
101             while (t != nill){
102                   splay(t);
103                   if (p != nill) p->parent = t;
104                   t->ch[1] = p;
105
106                   p = t;
107                   t = t->parent;
108             }
109             return p;
110        }
111        //合并u,v所在的树
112        void merge(Node *u, Node *v){
113             access(u);
114             splay(u);
115             u->delta = 1;
116             u->parent = v;
117             return;
118        }
119        void cut(Node *u, Node *v){
120             access(u)->delta ^= 1;
121             //splay(u);//转到顶上就切断v就可以了
122             access(v);
123             splay(v);
124
125             v->ch[0]->parent = nill;
126             v->ch[0] = nill;
127             return;
128        }
129        bool check(Node *u, Node *v){
130             while (u->parent != nill) u = u->parent;
131             while (v->parent != nill) v = v->parent;
132             return (u == v);
133        }
134 }splay;
135 int n, m;
136
137 void init(){
138      splay.init();
139      scanf("%d%d", &n, &m);
140      for (int i = 1; i <= n; i++){
141          splay.NEW(i);
142      }
143 }
144 void work(){
145      for (int i = 1; i <= m; i++){
146          char str[10];
147          scanf("%s", str);
148          if (str[0] == 'C'){
149             int u, v;
150             scanf("%d%d", &u, &v);
151             if (u == v || splay.check(splay.node + u, splay.node + v)) continue;//已经在一个集合里面了
152             splay.merge(splay.node + u, splay.node + v);
153          }else if (str[0] == 'D'){
154              int u, v;
155              scanf("%d%d", &u, &v);
156              if (u == v || !splay.check(splay.node + u, splay.node + v)) continue;
157              splay.cut(splay.node + u, splay.node + v);
158          }else {
159              int u, v;
160              scanf("%d%d", &u, &v);
161              if (u == v || splay.check(splay.node + u, splay.node + v)) printf("Yes\n");
162              else printf("No\n");
163          }
164      }
165 }
166
167 int main (){
168
169     init();
170     work();
171     return 0;
172 }
View Code

posted @ 2015-03-16 20:15  TCtower  阅读(...)  评论(... 编辑 收藏