Can you answer these queries VII
Time Limit: 1000ms
Memory Limit: 262144KB
This problem will be judged on SPOJ. Original ID: GSS7
64-bit integer IO format: %lld Java class name: Main
Given a tree with N ( N<=100000 ) nodes. Each node has a interger value x_i ( |x_i|<=10000 ).
You have to apply Q ( Q<=100000 ) operations:
1. 1 a b : answer the maximum contiguous sum (maybe empty,will always larger than or equal to 0 ) from the path a->b ( inclusive ).
2. 2 a b c : change all value in the path a->b ( inclusive ) to c.
Input
first line consists one interger N.
next line consists N interger x_i.
next N-1 line , each consists two interger u,v , means that node u and node v are connected
next line consists 1 interger Q.
next Q line : 1 a b or 2 a b c .
Output
For each query, output one line the maximum contiguous sum.
Example
Input: 5
-3 -2 1 2 3
1 2
2 3
1 4
4 5
3
1 2 5
2 3 4 2
1 2 5 Output: 5
9
Source
解题：LCT
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int maxn = 100010;
4 struct arc{
5 int to,next;
6 arc(int x = 0,int y = -1){
7 to = x;
8 next = y;
9 }
10 }e[maxn<<1];
11 struct LCT{
12 int fa[maxn],ch[maxn][2],sz[maxn],parent[maxn];
13 int key[maxn],ls[maxn],rs[maxn],ms[maxn],sum[maxn];
14 bool reset[maxn];
15 inline void pushup(int x){
16 sz[x] = 1 + sz[ch[x][0]] + sz[ch[x][1]];
17 sum[x] = key[x] + sum[ch[x][0]] + sum[ch[x][1]];
18 ls[x] = max(ls[ch[x][0]],sum[ch[x][0]] + key[x] + ls[ch[x][1]]);
19 rs[x] = max(rs[ch[x][1]],sum[ch[x][1]] + key[x] + rs[ch[x][0]]);
20 ms[x] = max(ms[ch[x][0]],ms[ch[x][1]]);
21 ms[x] = max(ms[x],key[x] + rs[ch[x][0]] + ls[ch[x][1]]);
22 }
23 inline void set(int x,int val){
24 if(!x) return;
25 reset[x] = true;
26 key[x] = val;
27 sum[x] = sz[x]*val;
28 ls[x] = rs[x] = ms[x] = max(0,sum[x]);
29 }
30 inline void pushdown(int x){
31 if(reset[x]){
32 set(ch[x][0],key[x]);
33 set(ch[x][1],key[x]);
34 reset[x] = false;
35 }
36 }
37 void rotate(int x,int kd){
38 int y = fa[x];
39 pushdown(y);
40 pushdown(x);
41 ch[y][kd^1] = ch[x][kd];
42 fa[ch[x][kd]] = y;
43 fa[x] = fa[y];
44 ch[x][kd] = y;
45 fa[y] = x;
46 if(fa[x]) ch[fa[x]][y == ch[fa[x]][1]] = x;
47 pushup(y);
48 }
49 void splay(int x,int goal = 0){
50 pushdown(x);
51 int y = x;
52 while(fa[y]) y = fa[y];
53 if(x != y){
54 parent[x] = parent[y];
55 parent[y] = 0;
56 while(fa[x] != goal){
57 pushdown(fa[fa[x]]);
58 pushdown(fa[x]);
59 pushdown(x);
60 if(fa[fa[x]] == goal) rotate(x,x == ch[fa[x]][0]);
61 else{
62 int y = fa[x],z = fa[y],s = (ch[z][0] == y);
63 if(x == ch[y][s]){
64 rotate(x,s^1);
65 rotate(x,s);
66 }else{
67 rotate(y,s);
68 rotate(x,s);
69 }
70 }
71 }
72 pushup(x);
73 }
74 }
75 void access(int x){
76 for(int y = 0; x; x = parent[x]){
77 splay(x);
78 fa[ch[x][1]] = 0;
79 parent[ch[x][1]] = x;
80 ch[x][1] = y;
81 fa[y] = x;
82 parent[y] = 0;
83 y = x;
84 pushup(x);
85 }
86 }
87 void update(int x,int y,int z){
88 access(y);
89 for(y = 0; x; x = parent[x]){
90 splay(x);
91 if(!parent[x]){
92 key[x] = z;
93 set(y,z);
94 set(ch[x][1],z);
95 return;
96 }
97 fa[ch[x][1]] = 0;
98 parent[ch[x][1]] = x;
99 ch[x][1] = y;
100 fa[y] = x;
101 parent[y] = 0;
102 y = x;
103 pushup(x);
104 }
105 }
106 int query(int x,int y){
107 access(y);
108 for(y = 0; x; x = parent[x]){
109 splay(x);
110 if(!parent[x]){
111 int ret = max(ms[y],ms[ch[x][1]]);
112 ret = max(ret,key[x] + ls[y] + ls[ch[x][1]]);
113 return ret;
114 }
115 fa[ch[x][1]] = 0;
116 parent[ch[x][1]] = x;
117 ch[x][1] = y;
118 fa[y] = x;
119 parent[y] = 0;
120 y = x;
121 pushup(x);
122 }
123 }
124 void init(){
125 memset(fa,0,sizeof fa);
126 memset(parent,0,sizeof parent);
127 memset(reset,false,sizeof reset);
128 memset(ch,0,sizeof ch);
129 sz[0] = 0;
130 }
131 }lct;
132 int head[maxn],tot;
133 void add(int u,int v){
134 e[tot] = arc(v,head[u]);
135 head[u] = tot++;
136 }
137 void dfs(int u,int fa){
138 for(int i = head[u]; ~i; i = e[i].next){
139 if(e[i].to == fa) continue;
140 lct.parent[e[i].to] = u;
141 dfs(e[i].to,u);
142 }
143 }
144 int main(){
145 int n,m,u,v,x,y,z,op;
146 while(~scanf("%d",&n)){
147 memset(head,-1,sizeof head);
148 tot = 0;
149 lct.init();
150 for(int i = 1; i <= n; ++i){
151 lct.sz[i] = 1;
152 scanf("%d",&lct.key[i]);
153 lct.sum[i] = lct.key[i];
154 lct.ls[i] = lct.rs[i] = lct.ms[i] = max(0,lct.key[i]);
155 }
156 for(int i = 1; i < n; ++i){
157 scanf("%d%d",&u,&v);
158 add(u,v);
159 add(v,u);
160 }
161 dfs(1,1);
162 scanf("%d",&m);
163 while(m--){
164 scanf("%d%d%d",&op,&x,&y);
165 if(op == 1) printf("%d\n",lct.query(x,y));
166 else{
167 scanf("%d\n",&z);
168 lct.update(x,y,z);
169 }
170 }
171 }
172 return 0;
173 }
时间： 2024-08-10 23:12:31