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 + 10;
vector<int> vt[MAXN];
int dp[MAXN << 1][30];
int first[MAXN], ver[MAXN << 1], deep[MAXN << 1];
int pre[MAXN], val[MAXN], yy[MAXN], ip = 0, indx = 0;

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 = 0; i < vt[u].size(); i++){
        int v = vt[u][i];
        if(v != fa){
            dfs(v, h + 1, u);
            ver[++indx] = u;
            deep[indx] = h;
        }
    }
}

void ST(int n){
    for(int i = 1; i <= n; i++){
        dp[i][0] = i;
    }
    for(int j = 1; (1 << j) <= n; j++){
        for(int i = 1; i + (1 << j) - 1 <= n; i++){
            int x = dp[i][j - 1], y = dp[i + (1 << (j - 1))][j -1];
            dp[i][j] = deep[x] < deep[y] ? x : y;
        }
    }
}

int RMQ(int l, int r){
    int len = log2(r - l + 1);
    int x = dp[l][len], y = dp[r - (1 << len) + 1][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 != -1){
        yy[pos++] = val[x];
        x = pre[x];
    }
}

void solve(int x, int y, int k){
    int pos = 0, 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
", yy[k - 1]);
    }
}

int main(void){
    int n, q, x, y, op;
    scanf("%d%d", &n, &q);
    for(int i = 1; i <= n; i++){
        scanf("%d", &val[i]);
    }
    for(int i = 1; i < n; i++){
        scanf("%d%d", &x, &y);
        vt[x].push_back(y);
        vt[y].push_back(x);
    }
    dfs(1, 1, -1);
    ST(indx);
    while(q--){
        scanf("%d%d%d", &op, &x, &y);
        if(!op) val[x] = y;
        else solve(x, y, op);
    }
    return 0;
}