HDU 3078 Network - 码农教程

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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题目大意：

给出一棵有n个节点的无根树

给出每个节点的权值

有m次询问

每次三个数x,y,z

若x==0 将y位置的数修改为z

若x！=0 询问y节点到z节点的权值总和

这道题数据规模比较小&&HDOJ的评测机速度还算快，求出LCA打暴力其实可以水过

当然如果你不介意的话可以用树链剖分+线段树+平衡树&&二分

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<algorithm>
  4 using namespace std;
  5 const int MAXN=2*1e5+10;
  6 inline int read()
  7 {
  8     char c=getchar();int x=0,f=1;
  9     while(c<'0'||c>'9')    {if(c=='-')    f=-1;c=getchar();}
 10     while(c>='0'&&c<='9')    x=x*10+c-48,c=getchar();return x*f;
 11 }
 12 int n,m,S=1; 
 13 int f[MAXN][21],deep[MAXN],g[MAXN];
 14 struct node
 15 {
 16     int u,v,w,nxt;
 17 }edge[MAXN];
 18 int head[MAXN];
 19 int num=1;
 20 int a[MAXN];
 21 inline void add_edge(int x,int y,int z)
 22 {
 23     edge[num].u=x;
 24     edge[num].v=y;
 25     edge[num].w=z;
 26     edge[num].nxt=head[x];
 27     head[x]=num++;
 28 }
 29 void dfs(int now)
 30 {
 31     for(int i=head[now];i!=-1;i=edge[i].nxt)
 32         if(deep[edge[i].v]==0)
 33         {
 34             deep[edge[i].v]=deep[now]+1;
 35             f[edge[i].v][0]=now;
 36             g[edge[i].v]=g[now]+edge[i].w;
 37             dfs(edge[i].v);
 38         }
 39             
 40 }
 41 inline void pre()
 42 {
 43     for(int i=1;i<=19;i++)
 44         for(int j=1;j<=n;j++)
 45             f[j][i]=f[f[j][i-1]][i-1];
 46 }
 47 inline int LCA(int x,int y)
 48 {
 49     if(deep[x]<deep[y])    swap(x,y);
 50     for(int i=19;i>=0;i--)
 51         if(deep[f[x][i]]>=deep[y])
 52             x=f[x][i];
 53     if(x==y)    return x;
 54 
 55     for(int i=19;i>=0;i--)
 56         if(f[x][i]!=f[y][i])
 57             x=f[x][i],y=f[y][i];
 58     return f[x][0];
 59 }
 60 int tot=0;
 61 int ans[MAXN];
 62 int comp(const int &a,const int &b)
 63 {
 64     return a>b;
 65 }
 66 int work(int k,int x,int y)
 67 {
 68     tot=0;
 69     int lca=LCA(x,y);
 70     while(x!=lca)    ans[++tot]=a[x],x=f[x][0];ans[++tot]=a[x];
 71     while(y!=lca)    ans[++tot]=a[y],y=f[y][0];
 72     sort(ans+1,ans+tot+1,comp);
 73     if(k>tot)    printf("invalid request!n");
 74     else         printf("%dn",ans[k]);
 75 }
 76 int main()
 77 {
 78     int T=1;
 79     while(T--)
 80     {
 81         n=read();m=read();
 82         for(int i=1;i<=n;i++)    a[i]=read();
 83         memset(head,-1,sizeof(head));num=1;
 84         memset(f,0,sizeof(f));
 85         memset(deep,0,sizeof(deep));
 86         for(int i=1;i<=n-1;i++)
 87         {
 88             int x=read(),y=read();
 89             add_edge(x,y,0);
 90             add_edge(y,x,0);
 91         }
 92         deep[S]=1;
 93         dfs(S);pre();
 94         while(m--)
 95         {
 96             
 97             int x=read(),y=read(),z=read();
 98             if(x==0)    a[y]=z;
 99             else work(x,y,z);
100         }    
101     }
102     
103     return 0;
104 }