SPOJ Query on a tree （树链剖分）_query on a tree spoj-优快云博客

本文链接：https://blog.youkuaiyun.com/acvcla/article/details/48508109

本文深入探讨了大数据开发领域的核心技术，包括Hadoop、Spark等分布式处理框架，以及Hive、Cassandra等数据存储解决方案。通过具体案例分析，展示了如何利用这些技术进行数据处理、分析和挖掘，为企业提供数据驱动的决策支持。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

Prev Submit Status Statistics Discuss Next

Font Size:

Type:

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

Tags ( Click to see )

Prev Submit Status Statistics Discuss Next

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;
}