Can you answer these queries VII
Time Limit: 1000ms
Memory Limit: 262144KB
This problem will be judged on SPOJ. Original ID: GSS764-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 ",lct.query(x,y)); 166 else{ 167 scanf("%d ",&z); 168 lct.update(x,y,z); 169 } 170 } 171 } 172 return 0; 173 }