Description
You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. We will ask you to perfrom some instructions of the following form:
CHANGE i ti : change the cost of the i-th edge to ti or
QUERY a b : ask for the maximum edge cost on the path from node a to node b
给一棵共有 n(n · 10000) 个结点的树, 每条边都有一个权值, 要求维护一个数据结构, 支持如下操作:
- 修改某条边的权值;
- 询问某两个结点之间的唯一通路上的最大边权.
其中操作的总次数为 q.
Input
The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.
For each test case:
In the first line there is an integer N (N <= 10000),
In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 1000000),
The next lines contain instructions "CHANGE i ti" or "QUERY a b",
The end of each test case is signified by the string "DONE".
There is one blank line between successive tests.
Output
For each "QUERY" operation, write one integer representing its result.
Sample Input
1
3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE
Sample Output
1
3
一道树链剖分的板子题，如果不会的话可以看浅谈算法——树链剖分。
稍微需要注意的就是要将边权全部都下放到点权上去，这样的话方便维护
/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline int read(){
int x=0,f=1;char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x>=10) print(x/10);
putchar(x%10+'0');
}
const int N=1e4;
int dfn[N+10],ID[N+10],V[N+10],line[N+10],n;
struct Segment{
#define ls (p<<1)
#define rs (p<<1|1)
int tree[(N<<2)+10];
void init(){memset(tree,128,sizeof(tree));}
void build(int p,int l,int r){
if (l==r){
tree[p]=V[dfn[l]];
return;
}
int mid=(l+r)>>1;
build(ls,l,mid),build(rs,mid+1,r);
tree[p]=max(tree[ls],tree[rs]);
}
void change(int p,int l,int r,int x,int v){
if (l==r){
tree[p]=v;
return;
}
int mid=(l+r)>>1;
if (x<=mid) change(ls,l,mid,x,v);
else change(rs,mid+1,r,x,v);
tree[p]=max(tree[ls],tree[rs]);
}
int Query(int p,int l,int r,int x,int y){
if (x<=l&&r<=y) return tree[p];
int mid=(l+r)>>1,res=0;
if (x<=mid) res=max(res,Query(ls,l,mid,x,y));
if (y>mid) res=max(res,Query(rs,mid+1,r,x,y));
return res;
}
}Tree;
struct S1{
int pre[(N<<1)+10],now[N+10],child[(N<<1)+10],val[(N<<1)+10],tot,cnt;
int fa[N+10],deep[N+10],top[N+10],size[N+10],Rem[N+10];
void join(int x,int y,int z){pre[++tot]=now[x],now[x]=tot,child[tot]=y,val[tot]=z;}
void insert(int x,int y,int z){join(x,y,z),join(y,x,z);}
void dfs(int x,int Deep){
int T=0; size[x]=1,deep[x]=Deep;
for (int p=now[x],son=child[p];p;p=pre[p],son=child[p]){
if (son==fa[x]) continue;
fa[son]=x,V[son]=val[p],line[(p+1)>>1]=son;
dfs(son,Deep+1);
size[x]+=size[son];
if (T<size[son]) T=size[son],Rem[x]=son;
}
}
void build(int x){
if (!x) return;
top[x]=Rem[fa[x]]==x?top[fa[x]]:x;
dfn[ID[x]=++cnt]=x;
build(Rem[x]);
for (int p=now[x],son=child[p];p;p=pre[p],son=child[p]){
if (son==fa[x]||son==Rem[x]) continue;
build(son);
}
}
int Query(int x,int y){
int res=-inf;
while (top[x]!=top[y]){
if (deep[top[x]]<deep[top[y]]) swap(x,y);
res=max(res,Tree.Query(1,1,n,ID[top[x]],ID[x]));
x=fa[top[x]];
}
if (x==y) return res;
if (deep[x]>deep[y]) swap(x,y);
res=max(res,Tree.Query(1,1,n,ID[Rem[x]],ID[y]));
return res;
}
void init(){
tot=cnt=0;
memset(fa,0,sizeof(fa));
memset(now,0,sizeof(now));
memset(Rem,0,sizeof(Rem));
}
}T;
void init(){
Tree.init();
T.init();
}
char s[10];
int main(){
for (int Data=read();Data;Data--){
init();
n=read();
for (int i=1;i<n;i++){
int x=read(),y=read(),z=read();
T.insert(x,y,z);
}
T.dfs(1,1),T.build(1),Tree.build(1,1,n);
while (true){
scanf("%s",s);
if (s[0]=='D') break;
int x=read(),y=read();
if (s[0]=='Q') printf("%d
",T.Query(x,y));
if (s[0]=='C') Tree.change(1,1,n,ID[line[x]],y);
}
}
return 0;
}