Problem Description
You are given a tree with N nodes which are numbered by integers 1..N. Each node is associated with an integer as the weight.
Your task is to deal with M operations of 4 types:
1.Delete an edge (x, y) from the tree, and then add a new edge (a, b). We ensure that it still constitutes a tree after adding the new edge.
2.Given two nodes a and b in the tree, change the weights of all the nodes on the path connecting node a and b (including node a and b) to a particular value x.
3.Given two nodes a and b in the tree, increase the weights of all the nodes on the path connecting node a and b (including node a and b) by a particular value d.
4.Given two nodes a and b in the tree, compute the second largest weight on the path connecting node a and b (including node a and b), and the number of times this weight occurs on the path. Note that here we need the strict second largest weight. For instance, the strict second largest weight of {3, 5, 2, 5, 3} is 3.
Input
The first line contains an integer T (T<=3), which means there are T test cases in the input.
For each test case, the first line contains two integers N and M (N, M<=10^5). The second line contains N integers, and the i-th integer is the weight of the i-th node in the tree (their absolute values are not larger than 10^4).
In next N-1 lines, there are two integers a and b (1<=a, b<=N), which means there exists an edge connecting node a and b.
The next M lines describe the operations you have to deal with. In each line the first integer is c (1<=c<=4), which indicates the type of operation.
If c = 1, there are four integers x, y, a, b (1<= x, y, a, b <=N) after c.
If c = 2, there are three integers a, b, x (1<= a, b<=N, |x|<=10^4) after c.
If c = 3, there are three integers a, b, d (1<= a, b<=N, |d|<=10^4) after c.
If c = 4 (it is a query operation), there are two integers a, b (1<= a, b<=N) after c.
All these parameters have the same meaning as described in problem description.
Output
For each test case, first output "Case #x:"" (x means case ID) in a separate line.
For each query operation, output two values: the second largest weight and the number of times it occurs. If the weights of nodes on that path are all the same, just output "ALL SAME" (without quotes).
题目大意：维护一棵树，每次删边加边、给一条路径的所有点赋一个值、给一条路径的所有点加上一个值，或询问一条路径上的第二大值及其在这条路径上的出现次数。
思路：算是LCT的模板题吧，维护每个区间的第一大值和第一大值的出现次数，第二大值和第二大值的出现次数。
PS：终于搞出了一个指针版，新模板出来啦~~~
犯过的错误：
1、LCT里splay的旋转和普通splay的旋转有所不同。
2、不能更新超级儿子 nil 的最值。
代码（2859MS）：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 using namespace std;
6 typedef long long LL;
7 #define FOR(i, n) for(int i = 0; i < n; ++i)
8
9 const int MAXV = 100010;
10 const int MAXE = MAXV << 1;
11 const int INF = 0x3f3f3f3f;
12 const int NINF = -INF;
13
14 struct LCT {
15 struct Node {
16 Node *ch[2], *fa;
17 int val, set, add;
18 int max[2], cnt[2], size;
19 bool rt, rev;
20 } statePool[MAXV], *nil;
21 int ncnt;
22
23 int head[MAXV], val[MAXV], ecnt;
24 int to[MAXE], next[MAXE];
25 int n, m, T;
26 Node *ptr[MAXV];
27
28 LCT() {
29 ptr[0] = nil = statePool;
30 nil->size = 0;
31 FOR(k, 2) nil->max[k] = NINF;
32 }
33
34 void init() {
35 memset(head + 1, -1, n * sizeof(int));
36 ncnt = 1;
37 ecnt = 0;
38 }
39
40 void add_edge(int u, int v) {
41 to[ecnt] = v; next[ecnt] = head[u]; head[u] = ecnt++;
42 to[ecnt] = u; next[ecnt] = head[v]; head[v] = ecnt++;
43 }
44
45 Node* new_node(int val, Node *f) {
46 Node* x = &statePool[ncnt++];
47 x->ch[0] = x->ch[1] = nil; x->fa = f;
48 x->val = val; x->set = NINF; x->add = 0;
49 x->max[0] = val; x->cnt[0] = 1;
50 x->max[1] = NINF;
51 x->size = 1;
52 x->rt = true; x->rev = false;
53 return x;
54 }
55
56 void dfs(int u, int f) {
57 ptr[u] = new_node(val[u], ptr[f]);
58 for(int p = head[u]; ~p; p = next[p]) {
59 int v = to[p];
60 if(v == f) continue;
61 dfs(v, u);
62 }
63 }
64
65 void get_max(int &a, int &b, int c) {
66 if(a != c) {
67 if(b < c) swap(b, c);
68 if(a < b) swap(a, b);
69 }
70 }
71
72 void cnt_max(int a, int &cnt, int b, int bcnt) {
73 if(a != NINF && a == b) cnt += bcnt;
74 }
75
76 void update(Node *x) {
77 x->size = x->ch[0]->size + x->ch[1]->size + 1;
78
79 x->max[0] = x->val; x->max[1] = NINF;
80 FOR(i, 2) FOR(j, 2)
81 get_max(x->max[0], x->max[1], x->ch[i]->max[j]);
82
83 FOR(k, 2) x->cnt[k] = 0;
84 FOR(k, 2) cnt_max(x->max[k], x->cnt[k], x->val, 1);
85 FOR(k, 2) FOR(i, 2) FOR(j, 2)
86 cnt_max(x->max[k], x->cnt[k], x->ch[i]->max[j], x->ch[i]->cnt[j]);
87 }
88
89 void rotate(Node *x) {
90 Node *y = x->fa;
91 int t = (y->ch[1] == x);
92
93 if(y->rt) y->rt = false, x->rt = true;
94 else y->fa->ch[y->fa->ch[1] == y] = x;
95 x->fa = y->fa;
96
97 (y->ch[t] = x->ch[t ^ 1])->fa = y;
98 (x->ch[t ^ 1] = y)->fa = x;
99 update(y);
100 }
101
102 void update_set(Node *x, int val) {
103 if(x == nil) return ;
104 x->add = 0;
105 x->val = x->set = val;
106 x->max[0] = val; x->cnt[0] = x->size;
107 x->max[1] = NINF;
108 }
109
110 void update_add(Node *x, int val) {
111 if(x == nil) return ;
112 x->add += val;
113 x->val += val;
114 FOR(k, 2) if(x->max[k] != NINF)
115 x->max[k] += val;
116 }
117
118 void update_rev(Node *x) {
119 if(x == nil) return ;
120 x->rev = !x->rev;
121 swap(x->ch[0], x->ch[1]);
122 }
123
124 void pushdown(Node *x) {
125 if(x->set != NINF) {
126 FOR(k, 2) update_set(x->ch[k], x->set);
127 x->set = NINF;
128 }
129 if(x->add != 0) {
130 FOR(k, 2) update_add(x->ch[k], x->add);
131 x->add = 0;
132 }
133 if(x->rev) {
134 FOR(k, 2) update_rev(x->ch[k]);
135 x->rev = false;
136 }
137 }
138
139 void push(Node *x) {
140 if(!x->rt) push(x->fa);
141 pushdown(x);
142 }
143
144 void splay(Node *x) {
145 push(x);
146 while(!x->rt) {
147 Node *f = x->fa, *ff = f->fa;
148 if(!f->rt) rotate(((ff->ch[1] == f) && (f->ch[1] == x)) ? f : x);
149 rotate(x);
150 }
151 update(x);
152 }
153
154 Node* access(Node *x) {
155 Node *y = nil;
156 while(x != nil) {
157 splay(x);
158 x->ch[1]->rt = true;
159 (x->ch[1] = y)->rt = false;
160 update(x);
161 y = x; x = x->fa;
162 }
163 return y;
164 }
165
166 void be_root(Node *x) {
167 access(x);
168 splay(x);
169 update_rev(x);
170 }
171
172 void link(Node *x, Node *y) {
173 be_root(x);
174 x->fa = y;
175 }
176
177 void cut(Node *x, Node *y) {
178 be_root(x);
179 access(x);
180 splay(y);
181 y->fa = nil;
182 }
183
184 void modify_add(Node *x, Node *y, int w) {
185 be_root(x);
186 update_add(access(y), w);
187 }
188
189 void modify_set(Node *x, Node *y, int w) {
190 be_root(x);
191 update_set(access(y), w);
192 }
193
194 void query(Node *x, Node *y) {
195 be_root(x);
196 Node *r = access(y);
197 if(r->max[1] == NINF) puts("ALL SAME");
198 else printf("%d %d\n", r->max[1], r->cnt[1]);
199 }
200
201 void work() {
202 scanf("%d", &T);
203 for(int t = 1; t <= T; ++t) {
204 scanf("%d%d", &n, &m);
205 init();
206 for(int i = 1; i <= n; ++i) scanf("%d", &val[i]);
207 for(int i = 1, u, v; i < n; ++i) {
208 scanf("%d%d", &u, &v);
209 add_edge(u, v);
210 }
211 dfs(1, 0);
212 printf("Case #%d:\n", t);
213 for(int i = 0, x, y, a, b, op; i < m; ++i) {
214 scanf("%d", &op);
215 if(op == 1) {
216 scanf("%d%d%d%d", &x, &y, &a, &b);
217 cut(ptr[x], ptr[y]);
218 link(ptr[a], ptr[b]);
219 } else if(op == 2) {
220 scanf("%d%d%d", &a, &b, &x);
221 modify_set(ptr[a], ptr[b], x);
222 } else if(op == 3) {
223 scanf("%d%d%d", &a, &b, &x);
224 modify_add(ptr[a], ptr[b], x);
225 } else {
226 scanf("%d%d", &a, &b);
227 query(ptr[a], ptr[b]);
228 }
229 }
230 }
231 }
232 } S;
233
234 int main() {
235 S.work();
236 }