Network （hdu 3078 在线LCA）

Network

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 683 Accepted Submission(s): 274



Problem Description

The ALPC company is now working on his own network system, which is connecting all N ALPC department. To economize on spending, the backbone network has only one router for each department, and N-1 optical fiber in total to connect all routers.

The usual way to measure connecting speed is lag, or network latency, referring the time taken for a sent packet of data to be received at the other end.

Now the network is on trial, and new photonic crystal fibers designed by ALPC42 is trying out, the lag on fibers can be ignored. That means, lag happened when message transport through the router. ALPC42 is trying to change routers to make the network faster,
now he want to know that, which router, in any exactly time, between any pair of nodes, the K-th high latency is. He needs your help.



Input

There are only one test case in input file.

Your program is able to get the information of N routers and N-1 fiber connections from input, and Q questions for two condition: 1. For some reason, the latency of one router changed. 2. Querying the K-th longest lag router between two routers.

For each data case, two integers N and Q for first line. 0<=N<=80000, 0<=Q<=30000.

Then n integers in second line refer to the latency of each router in the very beginning.

Then N-1 lines followed, contains two integers x and y for each, telling there is a fiber connect router x and router y.

Then q lines followed to describe questions, three numbers k, a, b for each line. If k=0, Telling the latency of router a, Ta changed to b; if k>0, asking the latency of the k-th longest lag router between a and b (include router a and b). 0<=b<100000000.

A blank line follows after each case.



Output

For each question k>0, print a line to answer the latency time. Once there are less than k routers in the way, print "invalid request!" instead.



Sample Input

5 5
5 1 2 3 4
3 1
2 1
4 3
5 3
2 4 5
0 1 2
2 2 3
2 1 4
3 3 5

Sample Output

3
2
2
invalid request!

Source

2009 Multi-University Training Contest 17 - Host
by NUDT



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题意：n个点n-1条边，每个点有一个权值value[i]，q次询问（k，u，v），若k==0，令value[u]=v；若k>0，输出u到v路径上第k大的权值。

思路：在线LCA算法，LCA转RMQ，添加一个pre数组记录路径。

代码：

#include <iostream>
#include <functional>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000009
const int maxn = 1005;
const int MAXN = 80010;
const int MAXM = 200010;

int n,q;

int rmq[2*MAXN];
int pre[MAXN];

struct ST{
    int mm[2*MAXN];
    int dp[2*MAXN][20];
    void init(int n)
    {
        mm[0]=-1;
        for (int i=1;i<=n;i++){
            mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
            dp[i][0]=i;
        }
        for (int j=1;j<=mm
;j++)
            for (int i=1;i+(1<<j)-1<=n;i++)
            dp[i][j]=rmq[dp[i][j-1]]<rmq[dp[i+(1<<(j-1))][j-1]]?dp[i][j-1]:dp[i+(1<<(j-1))][j-1];
    }
    int query(int a,int b)
    {
        if (a>b) swap(a,b);
        int k=mm[b-a+1];
        return rmq[dp[a][k]]<=rmq[dp[b-(1<<k)+1][k]]?dp[a][k]:dp[b-(1<<k)+1][k];
    }
};

struct Edge{
    int to,next;
}edge[MAXN*2];

int tot,head[MAXN];

int F[MAXN*2];
int P[MAXN];
int cnt;
ST st;

void init()
{
    tot=0;
    memset(head,-1,sizeof(head));
    memset(pre,-1,sizeof(pre));
}

void addedge(int u,int v)
{
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void dfs(int u,int father,int dep)
{
    F[++cnt]=u;
    rmq[cnt]=dep;
    P[u]=cnt;
    for (int i=head[u];~i;i=edge[i].next)
    {
        int v=edge[i].to;
        if (v==father) continue;
        pre[v]=u;
        dfs(v,u,dep+1);
        F[++cnt]=u;
        rmq[cnt]=dep;
    }
}

void LCA_init(int root,int node_num)
{
    cnt=0;
    dfs(root,root,0);
    st.init(2*node_num-1);
}

int query(int u,int v)
{
    return F[st.query(P[u],P[v])];
}

bool flag[MAXN];
int value[MAXN];

int ans[MAXN];

void solve(int u,int v,int p)
{
    int f=query(u,v);
    int cc=0;
    while (u!=f)
    {
        ans[cc++]=value[u];
        u=pre[u];
    }
    while (v!=f)
    {
        ans[cc++]=value[v];
        v=pre[v];
    }
    ans[cc++]=value[f];
    if (cc<p) printf("invalid request!\n");
    else
    {
        sort(ans,ans+cc);
        printf("%d\n",ans[cc-p]);
    }
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    int i,j,u,v,k;
    while (~scanf("%d%d",&n,&q))
    {
        for (i=1;i<=n;i++)
            scanf("%d",&value[i]);
        init();
        memset(flag,false,sizeof(flag));
        for (i=1;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
            flag[v]=true;
        }
        int root;
        for (i=1;i<=n;i++)
        {
            if (!flag[i])
            {
                root=i;
                break;
            }
        }
        LCA_init(root,n);
        while (q--)
        {
            scanf("%d%d%d",&k,&u,&v);
            if (k==0)
                value[u]=v;
            else solve(u,v,k);
        }
    }
    return 0;
}