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SPOJ QTREE - Query on a tree

时间:2017-08-24 21:38:10      阅读:211      评论:0      收藏:0      [点我收藏+]

标签:integer   space   uil   main   bit   const   nlogn   inf   put   

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

题意:给一棵树,边上有权值,多组询问,问任意两个路劲之间的最大边权,修改某一条边的权值。

分析:

线段树已经炉火纯青了,重新标号后,可以build操作,但是想着update更方便,也只是O(nlogn),差不了多少。

点权会容易操作的多,但是这里是边权,我就将线段树底层第一个点至为 -inf,然后按照顺序加上边的权值,边权就变成线段树里面的点权了。

#include <bits/stdc++.h>

using namespace std;

const int maxn=10005;
vector <pair<int,int> > g[maxn];
int num;
const int inf=1<<29;
int id[maxn],deep[maxn],siz[maxn],son[maxn],fa[maxn];
int val[maxn],top[maxn];

struct Edge
{
    int x,y,val;
    void read()
    {
        scanf("%d %d %d",&x,&y,&val);
        g[x].push_back(make_pair(y,val));
        g[y].push_back(make_pair(x,val));
    }
}edges[maxn+20];

void dfs1(int u,int father,int d,int va)
{
    deep[u]=d;
    siz[u]=1;
    son[u]=0;
    fa[u]=father;

    val[u] = va;

    for(int i=0;i<(int)g[u].size();i++)
    {
        int v=g[u][i].first;
        if(v==father)   continue;
        dfs1(v,u,d+1,g[u][i].second);
        siz[u]+=siz[v];
        if(siz[son[u]]<siz[v])  son[u]=v;
    }
}

void dfs2(int u,int tp)
{
    top[u]=tp;
    id[u]=++num;
    if(son[u])  dfs2(son[u],tp);
    for(int i=0;i<(int)g[u].size();i++)
    {
        int v=g[u][i].first;
        if(v==fa[u]||v==son[u]) continue;
        dfs2(v,v);
    }
}

int qL,qR;
int p,v;
int _max;
int n;
struct IntervalTree {
    int maxv[maxn*4];

    void build(int o,int L,int R) {
        int M = L + (R-L)/2;
        if(L==R) maxv[o] = val[L];
        else {
            build(o*2,L,M);
            build(o*2+1,M+1,R);
            maxv[o] = max(maxv[o*2],maxv[o*2+1]);
        }
    }

    void update(int o,int L,int R) {
        int M = L + (R-L)/2;
        if(L==R) maxv[o] = v;
        else {
            if(p<=M) update(o*2,L,M);
            else update(o*2+1,M+1,R);
            maxv[o] = max(maxv[o*2],maxv[o*2+1]);
        }
    }

    int qurey(int o,int L,int R) {
        int M = L + (R-L)/2, ans = -inf;
        if(qL<=L&&R<=qR) return maxv[o];
        if(qL<=M) ans = max(ans,qurey(o*2,L,M));
        if(M<qR) ans = max(ans,qurey(o*2+1,M+1,R));
        return ans;
    }

}sol;

int yougth(int u,int v) {
    int f1 = top[u],f2 = top[v];
    int ret = -inf;

    while(f1!=f2) {

        if(deep[f1]<deep[f2]) {
            swap(u,v);
            swap(f1,f2);
        }

        qL = id[f1],qR = id[u],_max = -inf;
        _max = max(_max,sol.qurey(1,1,n));
        ret = max(ret,_max);
        u = fa[f1];
        f1 = top[u];
    }

    if(u==v) return ret;
    if(deep[u]>deep[v]) swap(u,v);
    qL = id[son[u]],qR=id[v],_max = -inf;
    _max = max(_max,sol.qurey(1,1,n));
    ret = max(ret,_max);
    return ret;
}


int main(int argc, char const *argv[])
{
   // freopen("in.txt","r",stdin);
    int t;
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);

        for (int i = 0; i < maxn; ++i)
            g[i].clear();

        for (int i = 1; i < n; ++i)
            edges[i].read();

        num = 0;
        dfs1(1,0,1,-inf);
        dfs2(1,1);

        for(int i=1; i <=n; i++) {
            p = id[i];
            v = val[i];
            sol.update(1,1,n);
        }


        char cmd[20];
        while(scanf("%s",cmd),strcmp(cmd,"DONE")!=0) {
            if(strcmp(cmd,"QUERY")==0) {
                int u,v;
                scanf("%d%d",&u,&v);

                printf("%d\n",yougth(u,v));
            }
            else {
                int a,b;
                scanf("%d%d",&a,&b);
                int t = fa[edges[a].y] == edges[a].x ? edges[a].y : edges[a].x;
                p = id[t],v = b;
                sol.update(1,1,n);
            }
        }
    }
    return 0;
}

 

SPOJ QTREE - Query on a tree

标签:integer   space   uil   main   bit   const   nlogn   inf   put   

原文地址:http://www.cnblogs.com/TreeDream/p/7425130.html

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