Network(lca暴力)
Network
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/65536K (Java/Other)
Total Submission(s) : 24 Accepted Submission(s) : 5
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.
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.
latency time. Once there are less than k routers in the way, print "invalid
request!" instead.
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
2
2
invalid request!
NUDT
#include <iostream>
#include<cstring>
#include <string>
#include <algorithm>
using namespace std;
int f[];
int vis[];
int head[];
int dep[];
int qv[];
int cnt=;
int a[];
struct node
{
int v;
int nxt;
}e[*]; bool cmp(int x,int y)
{
return x>y;
} void add(int u,int v)
{
e[++cnt].nxt=head[u];
e[cnt].v = v;
head[u]=cnt;
return;
} void dfs(int u,int ff)
{
f[u]=ff;
dep[u]=dep[ff]+;
for(int i=head[u];i!=-;i=e[i].nxt)
{
if(e[i].v!=ff)//要防止重复搜索,tla了多次
{
dfs(e[i].v,u);
}
}
} int kk;
void lca(int x,int y)
{
//逻辑关系要理清
kk=;
if(dep[x]<dep[y])
{
swap(x,y);
}
while(dep[x]>dep[y])
{
a[kk++]=qv[x];
x=f[x];
}
if(x==y)
{
a[kk++]=x;
return;
}
if(f[x]==f[y])
{
a[kk++]=qv[x];
a[kk++]=qv[y];
a[kk++]=qv[f[x]];
return;
}
while(f[x]!=f[y])//直接暴力
{
a[kk++]=qv[x];
a[kk++]=qv[y];
x=f[x];
y=f[y];
}
a[kk++]=qv[x];
a[kk++]=qv[y];
a[kk++]=qv[f[x]];
return;
} int main()
{
int n,m;
scanf("%d %d",&n,&m);
memset(vis,,sizeof(vis));
memset(head,-,sizeof(head));
for(int i=;i<=n;i++)
{
scanf("%d",&qv[i]);
}
for(int i=;i<=n-;i++)
{
int u,v;
scanf("%d %d",&u,&v);
add(u,v);
add(v,u);
}
f[]=-;
dep[]=;
dfs(,);
for(int i=;i<=m;i++)
{
int u,v,p;
scanf("%d %d %d",&p,&u,&v);
if(p!=)
{ lca(u,v);
if(p>kk-) cout<<"invalid request!"<<endl;
else
{
sort(a+,a+kk,cmp);//要从大到小排
cout<<a[p]<<endl;
}
}
else qv[u]=v;
}
return ; }
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