QTREE - Query on a tree 树链剖分 或者 动态树
QTREE - Query on a tree
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
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.
Example
Input: 1 3 1 2 1 2 3 2 QUERY 1 2 CHANGE 1 3 QUERY 1 2 DONE Output: 1 3
我的样例
2
9
1 2 8
1 3 7
3 4 6
3 5 5
4 6 4
6 7 3
6 9 2
7 8 1
QUERY 1 2
QUERY 7 9
QUERY 8 2
QUERY 5 9
DONE
3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE
树链剖分，查询的时候不使用lca，而是利用辅助数组fa，两个儿子同时沿着自己的重链往上爬。
把边权转化为边上深度更大的结点的点权，因为不能根据直接指定线段树上某个点k的权值，所以建立线段树之后再逐个插入每个点k=pos[i]的点权val[i]。
#include
#include
#include
using namespace std;
#define inf 0x7fffffff
const int N = 10000 + 10;
int RD(){
// int x;
// scanf("%d",&x);
// return x;
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
struct Edge{int to,nxt,c;}e[N<<1];
int tot, head[N], n, sz, vis[N];
void initt(){
tot = 0;
for(int i=1;i<=n;i++) head[i] = vis[i] = 0;
sz = 0;
}
void add(int u,int v,int c){
e[++tot].to = v; e[tot].nxt = head[u]; e[tot].c = c; head[u] = tot;
e[++tot].to = u; e[tot].nxt = head[v]; e[tot].c = c; head[v] = tot;
}
int siz[N], fa[N], dep[N], val[N];
void dfs1(int u){
siz[u] = 1; vis[u] = 1;
for(int i=head[u];i;i=e[i].nxt){
int v = e[i].to;
if (vis[v]) continue;
dep[v] = dep[u] + 1;
val[v] = e[i].c;
fa[v] = u;
dfs1(v);
siz[u] += siz[v];
}
}
int pos[N], belong[N];
void dfs2(int u,int f){
int k = 0;
pos[u] = ++sz; belong[u] = f;
for(int i=head[u];i;i=e[i].nxt){
int v = e[i].to;
if (dep[v] > dep[u] && siz[v] > siz[k]) k = v;
}
if (!k) return;
dfs2(k, f);
for(int i=head[u];i;i=e[i].nxt){
int v = e[i].to;
if (dep[v] > dep[u] && v != k) dfs2(v, v);
}
}
struct Seg{int l,r,mx;}t[N<<2];
inline void pushup(int k){
t[k].mx = max(t[k<<1].mx, t[k<<1|1].mx);
}
void build(int k,int l,int r){
t[k].l = l; t[k].r = r;
t[k].mx = 0;
if (l==r) return;
int mid = (l+r)>>1;
build(k<<1, l, mid);
build(k<<1|1,mid+1,r);
}
inline void change(int k,int x,int y){
int l = t[k].l, r = t[k].r, mid = (l+r)>>1;
if (l == r){ t[k].mx = y; return; }
if (x <= mid) change(k<<1, x, y);
else change(k<<1|1, x, y);
pushup(k);
}
inline int queryMax(int k,int x,int y){
int l = t[k].l, r = t[k].r, mid = (l+r)>>1;
if (x <= l && r <= y) return t[k].mx;
if (y <= mid) return queryMax(k<<1, x, y);
else if (x > mid) return queryMax(k<<1|1, x, y);
else return max(queryMax(k<<1, x, mid), queryMax(k<<1|1, mid+1, y));
}
inline int solveMax(int x,int y){
int mx = -inf;
while(belong[x] != belong[y]){
if (dep[belong[x]] < dep[belong[y]]) swap(x,y);
mx = max(mx, queryMax(1, pos[belong[x]], pos[x]));
x = fa[belong[x]];
}
if (x == y) return mx;
if (dep[x] < dep[y]) swap(x, y);
if (pos[y]+1 <= pos[x]){
mx = max(mx, queryMax(1, pos[y]+1, pos[x]));
}
return mx;
}
int main()
{
// freopen("data.in", "r", stdin);
int T;
T = RD();
while(T--){
n = RD();
initt(); int a, b, c;
for(int i=1;i