hdu3078(lca / RMQ在线)

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3078

题意: 给出一棵 n 个点的带点权值的树, 接下来有 q 组形如 k, x, y 的输入, 若 k == 0 则将 x 点的权值替换成 y, 否则输出 x 到 y 之间顶点地 k 大的权值.

思路: 用一个数组 val 记录一下每个顶点的权值, 对于k == 0, 直接令 val[x] = y 即可 .

对于询问, 可以先求出 lca, 再记录一下路径上的顶点的权值, sort 一下, 输出第 k 大的即可.

代码:

 #include <iostream>
 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <algorithm>
 #include <vector>
 using namespace std;
 
 const int MAXN = 8e4 + ;
 vector<int> vt[MAXN];
 int dp[MAXN << ][];
 int first[MAXN], ver[MAXN << ], deep[MAXN << ];
 int pre[MAXN], val[MAXN], yy[MAXN], ip = , indx = ;
 
 bool cmp(int x, int y){
     return x > y;
 }
 
 void dfs(int u, int h, int fa){
     pre[u] = fa;
     ver[++indx] = u;
     deep[indx] = h;
     first[u] = indx;
     for(int i = ; i < vt[u].size(); i++){
         int v = vt[u][i];
         if(v != fa){
             dfs(v, h + , u);
             ver[++indx] = u;
             deep[indx] = h;
         }
     }
 }
 
 void ST(int n){
     for(int i = ; i <= n; i++){
         dp[i][] = i;
     }
     for(int j = ; ( << j) <= n; j++){
         for(int i = ; i + ( << j) -  <= n; i++){
             int x = dp[i][j - ], y = dp[i + ( << (j - ))][j -];
             dp[i][j] = deep[x] < deep[y] ? x : y;
         }
     }
 }
 
 int RMQ(int l, int r){
     int len = log2(r - l + );
     int x = dp[l][len], y = dp[r - ( << len) + ][len];
     return deep[x] < deep[y] ? x : y;
 }
 
 int LCA(int x, int y){
     int l = first[x];
     int r = first[y];
     if(l > r) swap(l, r);
     int pos = RMQ(l, r);
     return ver[pos];
 }
 
 void path(int x, int root, int &pos){
     while(x != root && x != -){
         yy[pos++] = val[x];
         x = pre[x];
     }
 }
 
 void solve(int x, int y, int k){
     int pos = , lca = LCA(x, y);
     path(x, lca, pos);
     path(y, lca, pos);
     yy[pos++] = val[lca];
     if(pos < k) puts("invalid request!");
     else{
         sort(yy, yy + pos, cmp);//注意是从大到小的第 k 大!!!!!!!!!
         printf("%d\n", yy[k - ]);
     }
 }
 
 int main(void){
     int n, q, x, y, op;
     scanf("%d%d", &n, &q);
     for(int i = ; i <= n; i++){
         scanf("%d", &val[i]);
     }
     for(int i = ; i < n; i++){
         scanf("%d%d", &x, &y);
         vt[x].push_back(y);
         vt[y].push_back(x);
     }
     dfs(, , -);
     ST(indx);
     while(q--){
         scanf("%d%d%d", &op, &x, &y);
         if(!op) val[x] = y;
         else solve(x, y, op);
     }
     return ;
 }