SPOJ Query on a tree （树链剖分）

Query on a tree

Time Limit: 5000ms
Memory Limit: 262144KB

This problem will be judged on SPOJ. Original ID: QTREE

64-bit integer IO format: %lld      Java class name: Main

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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

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代码:

#include <bits/stdc++.h>
#define lc(o) (o<<1)
#define rc(o) (o<<1|1)
using namespace std;
const int maxn = 1e4 + 10;

struct Edge{
int from,to,dist;
Edge(int u,int v,int w):from(u),to(v),dist(w){}
};

vector<int> g[maxn];
vector<Edge> edges;

void AddEdge(int u,int v,int w)
{
int m = edges.size();
g[u].push_back(m);
g[v].push_back(m);
edges.push_back(Edge(u,v,w));
}

int siz[maxn],son[maxn],fa[maxn],dep[maxn];
void pre(int u,int p)
{
dep[u] = dep[p] + 1;
siz[u] = 1;
fa[u] = p;
son[u] = 0;
for(vector<int>::iterator it = g[u].begin(); it != g[u].end(); ++it) {
Edge & e = edges[*it];
int v = e.to;
if(v == u) v = e.from;
if(v==p)continue;
pre(v,u);
siz[u] += siz[v];
if(siz[son[u]] < siz[v]) son[u] = v;
}
}

int root[maxn],pos[maxn],seg[maxn<<2],tot;

void built(int u,int t)
{
root[u] = t;
pos[u] = ++tot;
if(siz[u] == 1) return;
built(son[u],t);
for(vector<int>::iterator it = g[u].begin(); it != g[u].end(); ++it) {
Edge & e = edges[*it];
int v = e.to;
if(v == u) v = e.from;
if(v==fa[u]||v==son[u])continue;
built(v,v);
}
}
int ql,qr,val;
void Modify(int o,int L,int R)
{
if(L==R) {
seg[o] = val;
return;
}
int M = (L+R)>>1;
if(ql<=M) Modify(lc(o),L,M);
else Modify(rc(o),M+1,R);
seg[o] = max(seg[lc(o)],seg[rc(o)]);
}
int Query(int o,int L,int R)
{
if(ql<=L&&qr>=R) return seg[o];
int M = (L+R)>>1;
int ans = 0;
if(ql <= M) ans = Query(lc(o),L,M);
if(qr > M) ans = max(ans,Query(rc(o),M+1,R));
return ans;
}
int ask(int u,int v)
{
int ans = 0;
while(root[u] != root[v]) {
if(dep[root[u]] < dep[root[v]]) swap(u,v);
ql = pos[root[u]];
qr = pos[u];
if(ql <= qr) ans = max(ans,Query(1,1,tot));
u = fa[root[u]];
}
if(u==v)return ans;
ql = pos[u];
qr = pos[v];
if(ql > qr) swap(ql,qr);
ans = max(ans,Query(1,1,tot));
return ans;
}
int main()
{
int T;scanf("%d",&T);
siz[0] = 0;
while(T--) {
int n;scanf("%d",&n);
edges.clear();
edges.push_back(Edge(0,0,0));
for(int i = 1; i <= n; ++i) g[i].clear();
for(int i = 1; i < n; ++i) {
int u,v,w;scanf("%d%d%d",&u,&v,&w);
AddEdge(u,v,w);
}
pre(1,0);
tot = 0;
built(1,1);
memset(seg,0,sizeof(*seg)*(4*n));
for(int i = 1; i < n; ++i) {
Edge & e = edges[i-1];
if(dep[e.from] > dep[e.to]) swap(e.from,e.to);
val = e.dist;
ql = e.to;
Modify(1,1,tot);
}
char op[10];
int u,v;
while(scanf("%s",op) == 1)
{
if(op[0] == 'D')break;
scanf("%d%d",&u,&v);
if(op[0] == 'Q')
printf("%d\n",ask(u,v));
else {
ql = pos[edges[u-1].to];
val = v;
Modify(1,1,tot);
}
}
}
return 0;
}