LCA最近公共祖先

Posted 2021-03-03 guaguastandup

LCA

模板

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 100;
int head[maxn], to[maxn * 2], nxt[maxn * 2], d[maxn * 2], tot;
int n, m;
void add(int x, int y, int w){
    to[++tot] = y; nxt[tot] = head[x]; d[tot] = w; head[x] = tot;
}
int dp[maxn][20], dep[maxn], dis[maxn];
void dfs(int u, int fa){
    dp[u][0] = fa; dep[u] = dep[fa] + 1;
    for (int i = head[u]; i; i = nxt[i]){
        int v = to[i];
        if (v == fa) continue;
        dis[v] = dis[u] + d[i];
        dfs(v, u);
    }
}
void init(){
    memset(dp, 0, sizeof(dp));
    dep[0] = dis[0] = 0;
    dfs(1, 0); 
    for (int j = 1; j < 20; j++) 
        for (int i = 1; i <= n; i++) 
            dp[i][j] = dp[dp[i][j - 1]][j - 1];
}
int qlca(int x, int y){
    if (dep[x] < dep[y]) swap(x, y);
    int tmp = dep[x] - dep[y];
    for (int i = 0; tmp; i++, tmp >>= 1)
        if (tmp & 1) x = dp[x][i];
    if (x == y) return x;
    for (int i = 19; i >= 0; i--){
        if (dp[x][i] != dp[y][i]){
            x = dp[x][i]; y = dp[y][i];
        }
    }
    return dp[x][0];
}
int dist(int x,int y) {
    int u = qlca(x, y);
    int ans =  dis[x] + dis[y] - 2*dis[u];
    return dis[x] + dis[y] - 2*dis[u];
}

例题

[CF1304E]1-Trees and Queries

https://codeforces.com/contest/1304/problem/E

题意

给一棵树,每次查询时在x和y之间加一条边,然后问在a和b之间是否存在一条可以重复走的路径且这条路径的长度为k

思路

如果不走x和y的那条连边,那么a到b之间的路径长度为a与b的最短路径+2i,因为重复走的边对答案的贡献一定是偶数长度,只有走了x到y的那条连边,贡献为1,可以改变路径的奇偶性,枚举两种情况即可

代码 : https://xlorpaste.cn/nmnq8r