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.
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.
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
Recommend
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
", ans[i]);
else
printf("invalid request!
");
if(ans[rank])
printf("%d
", ans[rank]);
else
printf("invalid request!
");
return 0;
}