Network

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1293    Accepted Submission(s): 575

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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找树上两个点路径上的第k大节点

根据lca分别从起点和终点网上找点权，记下来，排序

lca预处理nlogn  lca查询logn  找路径 logn(n个节点的树最多logn层) 排序logn * logn

因此总复杂度(m + n)log^2n

不会超时

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
inline void read(int &x){char ch = getchar();char c = ch;x = 0;while(ch < ‘0‘ || ch > ‘9‘)c = ch, ch = getchar();while(ch <= ‘9‘ && ch >= ‘0‘)x = x * 10 + ch - ‘0‘, ch = getchar();if(c == ‘-‘)x = -x;}
inline void swap(int &a, int &b){int tmp = a;a = b;b = tmp;}

const int MAXN = 80000 + 10;

struct Edge{int u,v,next;}edge[MAXN << 1];
int head[MAXN],cnt,n,m,w[MAXN],p[20][MAXN],deep[MAXN],fa[MAXN],num[MAXN],b[MAXN];
inline void insert(int a,int b){edge[++cnt] = Edge{a, b, head[a]},head[a] = cnt;}

int llog2[MAXN],pow2[30];

void dfs(int u)
{
	for(int pos = head[u];pos;pos = edge[pos].next)
	{
		int v = edge[pos].v;
		if(b[v])continue;
		b[v] = true,deep[v] = deep[u] + 1,p[0][v] = u,fa[v] = u,dfs(v);
	}
}

inline void yuchuli()
{
	b[1] = true,deep[1] = 0,dfs(1);
	register int i;
	for(i = 1;i <= llog2[n];i ++)
		for(int j = n;j >= 1;j --)
			p[i][j] = p[i - 1][p[i - 1][j]];
}

inline int lca(int va, int vb)
{
	register int i;
	if(deep[va] < deep[vb])swap(va, vb);
	for(i = llog2[n];i >= 0;i --)
		if(deep[va] - pow2[i] >= deep[vb])
			va = p[i][va];
	if(va == vb)return va;
	for(i = llog2[n];i >= 0;i --)
		if(p[i][va] != p[i][vb])
			va = p[i][va],vb = p[i][vb];
	return p[0][va];
}

int ans[MAXN],rank;

inline void work(int s, int t, int k)
{
	int anc = lca(s, t);
	int now = s;
	register int i = 1;
	num[i] = w[now];
	while(now != anc)
		num[++i] = w[(now = fa[now])];
	now = t;
	while(now != anc)
		num[++i] = w[now], now = fa[now];
	std::sort(num + 1, num + 1 + i);
	if(i >= k)
		ans[++rank] = num[i - k + 1];
	else rank++;
}

int main()
{
	register int i,tmp1,tmp2,tmp3;
	llog2[0] = -1,pow2[0] = 1;
	for(int i = 1;i <= MAXN;i ++)llog2[i] = llog2[i >> 1] + 1;
	for(int i = 1;i <= 25; i++)pow2[i] = (pow2[i - 1] << 1);
	read(n),read(m);
	for(i = 1;i <= n;i ++)read(w[i]);
	for(i = 1;i < n;i ++)read(tmp1),read(tmp2),insert(tmp1, tmp2),insert(tmp2, tmp1);
	yuchuli();
	for(i = 1;i <= m;i ++)
	{
		read(tmp1),read(tmp2),read(tmp3);
		if(tmp1)work(tmp2, tmp3, tmp1);
		else w[tmp2] = tmp3;
	}
	for(int i = 1;i < rank;i ++)
		if(ans[i])
			printf("%d\n", ans[i]);
		else
			printf("invalid request!\n");
	if(ans[rank])
		printf("%d\n", ans[rank]);
	else
		printf("invalid request!\n");
	return 0;
}