Tree
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 7268 | Accepted: 1969 |
Description
You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edges are numbered 1 through N − 1. Each edge is associated with a weight. Then you are to execute a series of instructions on the tree. The instructions can be one of the following forms:
CHANGE i v |
Change the weight of the ith edge to v |
NEGATE a b |
Negate the weight of every edge on the path from a to b |
QUERY a b |
Find the maximum weight of edges on the path from a to b |
Input
The input contains multiple test cases. The first line of input contains an integer t (t ≤ 20), the number of test cases. Then follow the test cases.
Each test case is preceded by an empty line. The first nonempty line of its contains N (N ≤ 10,000). The next N − 1 lines each contains three integers a, b and c, describing an edge connecting nodes a and b with weight c. The edges are numbered in the order they appear in the input. Below them are the instructions, each sticking to the specification above. A lines with the word “DONE” ends the test case.
Output
For each “QUERY” instruction, output the result on a separate line.
Sample Input
1 3 1 2 1 2 3 2 QUERY 1 2 CHANGE 1 3 QUERY 1 2 DONE
Sample Output
1 3
Source
POJ Monthly--2007.06.03, Lei, Tao
【思路】
树链剖分，线段树。
又练了一边线段树<_<，bug满天飞T^T，线段树还得再做点题。
线段树：维护mx mn分别代表线段最大最小值，对于NEGATE操作打一个f标记，作用时直接将mx mn取负后交换即可。注意一下pushdown和maintain。
【代码】
1 #include<cstdio>
2 #include<cstring>
3 #include<vector>
4 #include<iostream>
5 #include<algorithm>
6 #define FOR(a,b,c) for(int a=(b);a<=(c);a++)
7 using namespace std;
8
9 const int N = 50000+10;
10 const int INF = 1e9;
11
12 struct Node { int l,r,mx,mn,f;
13 }T[N<<1];
14 struct Edge {
15 int u,v,w;
16 Edge(int u=0,int v=0,int w=0) :u(u),v(v),w(w) {};
17 };
18 int n,q,z,d[N][3];
19 vector<Edge> es;
20 vector<int> g[N];
21
22 void adde(int u,int v,int w) {
23 es.push_back(Edge(u,v,w));
24 int m=es.size();
25 g[u].push_back(m-1);
26 }
27 //INIT
28 int top[N],son[N],dep[N],fa[N],siz[N],w[N];
29 void dfs1(int u) {
30 son[u]=0; siz[u]=1;
31 for(int i=0;i<g[u].size();i++) {
32 int v=es[g[u][i]].v; //1
33 if(v!=fa[u]) {
34 fa[v]=u , dep[v]=dep[u]+1;
35 dfs1(v);
36 siz[u]+=siz[v];
37 if(siz[v]>siz[son[u]]) son[u]=v;
38 }
39 }
40 }
41 void dfs2(int u,int tp) {
42 top[u]=tp; w[u]=++z;
43 if(son[u]) dfs2(son[u],tp);
44 for(int i=0;i<g[u].size();i++) {
45 int v=es[g[u][i]].v;
46 if(v!=fa[u] && v!=son[u]) dfs2(v,v);
47 }
48 }
49 //SEGMENT TREE
50 //标记 该结点已被作用而子结点仍未作用
51 //pushdown 作用于访问节点 即只要访问就pushdown
52 void pushdown(int u) { //下传标记 同时作用于子结点
53 if(T[u].f && T[u].l<T[u].r) {
54 int lc=u<<1,rc=lc|1;
55 T[u].f=0;
56 T[lc].f^=1 ,T[rc].f^=1;
57 int t;
58 t=T[lc].mn,T[lc].mn=-T[lc].mx,T[lc].mx=-t;
59 t=T[rc].mn,T[rc].mn=-T[rc].mx,T[rc].mx=-t;
60 }
61 }
62 void maintain(int u) {
63 T[u].mx=max(T[u<<1].mx,T[u<<1|1].mx);
64 T[u].mn=min(T[u<<1].mn,T[u<<1|1].mn);
65 }
66 void build(int u,int L,int R) {
67 T[u].l=L,T[u].r=R;
68 T[u].mn=INF , T[u].mx=-INF , T[u].f=0;
69 if(L==R) return ; //2
70 int M=(L+R)>>1;
71 build(u<<1,L,M) , build(u<<1|1,M+1,R);
72 }
73 void change(int u,int r,int x) {
74 pushdown(u); //6
75 if(T[u].l==T[u].r) T[u].mn=T[u].mx=x;
76 else {
77 int M=(T[u].l+T[u].r)>>1; //3
78 if(r<=M) change(u<<1,r,x);
79 else change(u<<1|1,r,x);
80 maintain(u);
81 }
82 }
83 void Negate(int u,int L,int R) {
84 pushdown(u);
85 if(L<=T[u].l && T[u].r<=R) {//打标记 同时作用于当前结点
86 T[u].f=1;
87 int t; t=T[u].mn,T[u].mn=-T[u].mx,T[u].mx=-t; //4
88 }
89 else {
90 int M=(T[u].l+T[u].r)>>1;
91 if(L<=M) Negate(u<<1,L,R);
92 if(M<R) Negate(u<<1|1,L,R);
93 maintain(u);
94 }
95 }
96 int query(int u,int L,int R) {
97 pushdown(u);
98 if(L<=T[u].l && T[u].r<=R) return T[u].mx;
99 else {
100 int M=(T[u].l+T[u].r)>>1;
101 int ans=-INF;
102 if(L<=M) ans=max(ans,query(u<<1,L,R));
103 if(M<R) ans=max(ans,query(u<<1|1,L,R));
104 return ans;
105 }
106 }
107
108 //树链剖分
109 void modify(int u,int v) {
110 while(top[u]!=top[v]) {
111 if(dep[top[u]]<dep[top[v]]) swap(u,v);
112 Negate(1,w[top[u]],w[u]);
113 u=fa[top[u]];
114 }
115 if(u==v) return ;
116 if(dep[u]>dep[v]) swap(u,v);
117 Negate(1,w[son[u]],w[v]); //5
118 }
119 int query(int u,int v) {
120 int mx=-INF;
121 while(top[u]!=top[v]) {
122 if(dep[top[u]]<dep[top[v]]) swap(u,v);
123 mx=max(mx,query(1,w[top[u]],w[u]));
124 u=fa[top[u]];
125 }
126 if(u==v) return mx;
127 if(dep[u]>dep[v]) swap(u,v);
128 mx=max(mx,query(1,w[son[u]],w[v]));
129 return mx;
130 }
131
132 void read(int& x) {
133 char c=getchar();
134 while(!isdigit(c)) c=getchar();
135 x=0;
136 while(isdigit(c))
137 x=x*10+c-‘0‘ , c=getchar();
138 }
139 int main() {
140 //freopen("in.in","r",stdin);
141 //freopen("out.out","w",stdout);
142 int T; read(T);
143 while(T--) {
144 read(n);
145 z=0; es.clear();
146 FOR(i,0,n) g[i].clear();
147 int u,v,c;
148 FOR(i,1,n-1) {
149 read(u),read(v),read(c);
150 d[i][0]=u,d[i][1]=v,d[i][2]=c;
151 adde(u,v,c),adde(v,u,c);
152 }
153 dfs1(1),dfs2(1,1);
154 build(1,1,z);
155 FOR(i,1,n-1) {
156 if(dep[d[i][1]]<dep[d[i][0]]) swap(d[i][0],d[i][1]);
157 change(1,w[d[i][1]],d[i][2]);
158 }
159 char s[10];
160 while(scanf("%s",s)==1 && s[0]!=‘D‘) {
161 read(u),read(v);
162 if(s[0]==‘C‘) change(1,w[d[u][1]],v);
163 else if(s[0]==‘N‘) modify(u,v);
164 else printf("%d\n",query(u,v));
165 }
166 }
167 return 0;
168 }