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
题意：
对于一棵树，有几种操作：
Q ：x y 。问x到y之间的路径的最大边权值为多少。
C ：x y。把第x条边的权值改为y。
N：x y。把x到y之间的边权值取反。
D。结束。
思路：
和前面一道树剖题的查询是一样的，所以同样需要树剖+线段树，对于C操作，同样是线段树单点更新即可。但是需要区间权值取反，得用lazy标记一下。记录最大和最小值，方便在取反后还能得到最大值。
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define min(a,b) (a<b?a:b)
#define max(a,b) (a>b?a:b)
#define swap(a,b) (a^=b,b^=a,a^=b)
using namespace std;
const int maxn=200010;
int Laxt[maxn],Next[maxn],To[maxn],e[maxn][3],cnt;
int n,q; char opt[10];
struct TreeCut
{
int dpt[maxn],top[maxn],son[maxn],fa[maxn],sz[maxn],tot;
int tid[maxn],Rank[maxn],tim;
int Max[maxn<<2],Min[maxn<<2],Lazy[maxn<<2];
void init()
{
cnt=1; tim=0;
memset(Laxt,0,sizeof(Laxt));
memset(Max,0,sizeof(Max));
memset(Min,0,sizeof(Min));
memset(Lazy,0,sizeof(Lazy));
}
void add_edge(int u,int v)
{
Next[++cnt]=Laxt[u];
Laxt[u]=cnt; To[cnt]=v;
}
void dfs1(int u,int pre)
{
fa[u]=pre;dpt[u]=dpt[pre]+1;sz[u]=1;son[u]=0;
for(int i=Laxt[u];i;i=Next[i]){
int v=To[i]; if(v==pre) continue;
dfs1(v,u);sz[u]+=sz[v];
if(!son[u]||sz[v]>sz[son[u]]) son[u]=v;
}
}
void pushdown(int Now)
{
swap(Max[Now<<1],Min[Now<<1]),Max[Now<<1]=-Max[Now<<1],Min[Now<<1]=-Min[Now<<1];
swap(Max[Now<<1|1],Min[Now<<1|1]),Max[Now<<1|1]=-Max[Now<<1|1],Min[Now<<1|1]=-Min[Now<<1|1];
Lazy[Now<<1]^=1;Lazy[Now<<1|1]^=1;Lazy[Now]=0;
}
void pushup(int Now)
{
Max[Now]=max(Max[Now<<1],Max[Now<<1|1]);
Min[Now]=min(Min[Now<<1],Min[Now<<1|1]);
}
void dfs2(int u,int Top)
{
top[u]=Top; tid[u]=tim++;Rank[tid[u]]=u;
if(!son[u]) return ; dfs2(son[u],Top);
for(int i=Laxt[u];i;i=Next[i])
if(To[i]!=fa[u]&&To[i]!=son[u]) dfs2(To[i],To[i]);
}
void update(int Now,int L,int R,int pos,int val)
{
if(L==R){ Max[Now]=Min[Now]=val; return; }
if(Lazy[Now]&1) pushdown(Now);
int Mid=(L+R)>>1;
if(Mid>=pos) update(Now<<1,L,Mid,pos,val);
else update(Now<<1|1,Mid+1,R,pos,val);
pushup(Now);
}
int getmax(int Now,int L,int R,int l,int r)
{
if(L>=l&&R<=r) return Max[Now];
if(Lazy[Now]&1) pushdown(Now);
int Mid=(L+R)>>1,ans=-0x7fffffff;
if(r<=Mid) ans=getmax(Now<<1,L,Mid,l,r);
else if(l>Mid) ans=getmax(Now<<1|1,Mid+1,R,l,r);
else ans=max(getmax(Now<<1,L,Mid,l,Mid),getmax(Now<<1|1,Mid+1,R,Mid+1,r));
pushup(Now); return ans;
}
void addsign(int Now,int L,int R,int l,int r)
{
if(L>=l&&R<=r) {
Lazy[Now]^=1;swap(Max[Now],Min[Now]);
Max[Now]=-Max[Now];Min[Now]=-Min[Now];
return ;
}
if(Lazy[Now]&1) pushdown(Now);
int Mid=(L+R)>>1;
if(l<=Mid) addsign(Now<<1,L,Mid,l,r);
if(r>Mid) addsign(Now<<1|1,Mid+1,R,l,r);
pushup(Now);
}
void Make_Tree()
{
scanf("%d",&n);
for(int i=1;i<n;i++){
scanf("%d%d%d",&e[i][0],&e[i][1],&e[i][2]);
add_edge(e[i][0],e[i][1]);add_edge(e[i][1],e[i][0]);
} dfs1(1,0); dfs2(1,1);
for(int i=1;i<n;i++){
if(dpt[e[i][1]]<dpt[e[i][0]]) swap(e[i][1],e[i][0]);
update(1,1,n-1,tid[e[i][1]],e[i][2]);
}
}
int query(int u,int v)
{
int f1=top[u],f2=top[v],ans=-0x7fffffff;
while(f1!=f2){
if(dpt[f1]<dpt[f2]) swap(f1,f2),swap(u,v);
ans=max(ans,getmax(1,1,n-1,tid[f1],tid[u]));
u=fa[f1]; f1=top[u];
}
if(u!=v){
if(dpt[u]>dpt[v]) swap(u,v);
ans=max(ans,getmax(1,1,n-1,tid[son[u]],tid[v]));
} printf("%d\n",ans);
}
int Add_lazy(int u,int v)
{
int f1=top[u],f2=top[v];
while(f1!=f2){
if(dpt[f1]<dpt[f2]) swap(f1,f2),swap(u,v);
addsign(1,1,n-1,tid[f1],tid[u]);
u=fa[f1]; f1=top[u];
}
if(u!=v){
if(dpt[u]>dpt[v]) swap(u,v);
addsign(1,1,n-1,tid[son[u]],tid[v]);
}
}
void Query()
{
while(~scanf("%s",opt)) {
if(opt[0]=='D') return;
int x,y; scanf("%d%d",&x,&y);
if(opt[0]=='Q') query(x,y);
else if(opt[0]=='N') Add_lazy(x,y);
else update(1,1,n-1,tid[e[x][1]],y);
}
}
}Tc;
int main()
{
int T; scanf("%d",&T);
while(T--) {
Tc.init();
Tc.Make_Tree();
Tc.Query();
} return 0;
}
来源：https://www.cnblogs.com/hua-dong/p/8080872.html