Network LCA修改点权

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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一开始以为要选择第K大的这需要什么东西维护

结果就是排序  （打扰了）

 

题意：单case，一棵无根树，

输入点数和操作数，下面一行n个值代表每个点的权。下面n-1行是树边

操作分为

0 x w ，表示把点x的权改为w

k a b ， 求出，从a到b的路径中，第k大的点权

板子随便改一改就过了

 #include <cstdio>
 #include <cstring>
 #include <queue>
 #include <cmath>
 #include <algorithm>
 #include <set>
 #include <iostream>
 #include <map>
 #include <stack>
 #include <string>
 #include <vector>
 #define  pi acos(-1.0)
 #define  eps 1e-6
 #define  fi first
 #define  se second
 #define  lson l,m,rt<<1
 #define  rson m+1,r,rt<<1|1
 #define  bug         printf("******\n")
 #define  mem(a,b)    memset(a,b,sizeof(a))
 #define  fuck(x)     cout<<"["<<x<<"]"<<endl
 #define  f(a)        a*a
 #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  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  pf          printf
 #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  FIN         freopen("DATA.txt","r",stdin)
 #define  gcd(a,b)    __gcd(a,b)
 #define  lowbit(x)   x&-x
 #pragma  comment (linker,"/STACK:102400000,102400000")
 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 const int maxn = 2e5 + ;
 int _pow[maxn], dep[maxn], dis[maxn], vis[maxn], ver[maxn];
 int tot, head[maxn], dp[maxn * ][], k, first[maxn], path[maxn], val[maxn], fa[maxn];
 struct node {
     int u, v, w, nxt;
 } edge[maxn << ];
 void init() {
     tot = ;
     mem(head, -);
 }
 void add(int u, int v, int w) {
     edge[tot].v = v, edge[tot].u = u;
     edge[tot].w = w, edge[tot].nxt = head[u];
     head[u] = tot++;
 }
 void dfs(int u, int DEP, int pre) {
     vis[u] = ;
     ver[++k] = u;
     first[u] = k;
     dep[k] = DEP;
     fa[u] = pre;
     for (int i = head[u]; ~i; i = edge[i].nxt) {
         if (vis[edge[i].v]) continue;
         int v = edge[i].v, w = edge[i].w;
         dis[v] = dis[u] + w;
         dfs(v, DEP + , u);
         ver[++k] = u;
         dep[k] = DEP;
     }
 }
 void ST(int len) {
     int K = (int)(log((double)len) / log(2.0));
     for (int i =  ; i <= len ; i++) dp[i][] = i;
     for (int j =  ; j <= K ; j++) {
         for (int i =  ; i + _pow[j] -  <= len ; i++) {
             int a = dp[i][j - ], b = dp[i + _pow[j - ]][j - ];
             if (dep[a] < dep[b]) dp[i][j] = a;
             else dp[i][j] = b;
         }
     }
 }
 int RMQ(int x, int y) {
     int K = (int)(log((double)(y - x + )) / log(2.0));
     int a = dp[x][K], b = dp[y - _pow[K] + ][K];
     if (dep[a] < dep[b]) return a;
     else return b;
 }
 int LCA(int u, int v) {
     int x = first[u], y = first[v];
     if (x > y) swap(x, y);
     int ret = RMQ(x, y);
     return ver[ret];
 }
 int cnt;
 void solve(int s, int t) {
     while(s != t) {
         path[cnt++] = val[s];
         s = fa[s];
         // printf("cnt=%d\n",cnt);
     }
     path[cnt++] = val[t];
 }
 int cmp(int a, int b) {
     return a > b;
 }
 int main() {
     for (int i =  ; i <  ; i++) _pow[i] = ( << i);
     int n, m;
     while(~sff(n, m)) {
         init();
         mem(vis, );
         mem(fa, );
         for (int i =  ; i <= n ; i++) sf(val[i]);
         for (int i =  ; i < n -  ; i++) {
             int u, v;
             sff(u, v);
             add(u, v, );
             add(v, u, );
         }
         k = , dis[] = ;
         dfs(, , -);
         ST( * n - );
         while(m--) {
             int op, u, v;
             sfff(op,u,v);
             if (op == ) val[u] = v;
             else {
                 int lca = LCA(u, v);
                 //  printf("u=%d v=%d lca=%d\n",u,v,lca);
                 cnt = ;
                 solve(u, lca);
                 solve(v, lca);
                 cnt--;
               //  printf("cnt=%d\n", cnt);
                 if (op > cnt) printf("invalid request!\n");
                 else {
                     sort(path, path + cnt, cmp);
                     printf("%d\n", path[op - ]);
                 }
             }
         }
     }
     return  ;
 }