BZOJ4668: 冷战 (并查集 + LCA)

题意:动态给点连边 询问两个点之间最早是在第几个操作连起来的

题解:因为并查集按秩合并 秩最高是logn的 所以我们可以考虑把秩看作深度 跑LCA

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e5 + 5;

int n, m, cnt;
int fa[MAXN];
int id[MAXN];
int zhi[MAXN];

int find(int x) {
    if(fa[x]) return find(fa[x]);
    else return x;
}

int ffind(int x) {
    if(fa[x]) return ffind(fa[x]) + 1;
    else return 0;
}

void add(int x, int y, int z) {
    int fx = find(x);
    int fy = find(y);
    if(fx != fy) {
        if(zhi[fx] < zhi[fy]) {
            fa[fx] = fy; id[fx] = z;
        } else fa[fy] = fx, id[fy] = z;
        if(zhi[fx] == zhi[fy]) zhi[fx]++;
    }
}

int query(int x, int y) {
    int fx = find(x);
    int fy = find(y);
    if(fx != fy) return 0;
    int dx = ffind(x), dy = ffind(y);
    int res = 0;
    while(dx > dy) res = max(res, id[x]), x = fa[x], dx--;
    while(dx < dy) res = max(res, id[y]), y = fa[y], dy--;
    while(x != y) {
        res = max(res, max(id[x], id[y]));
        x = fa[x]; y = fa[y];
    }
    return res;
}

int main() {
    cnt = 0;
    scanf("%d%d", &n, &m);
    int las = 0;
    for(int i = 1; i <= m; i++) {
        int opt, u, v;
        scanf("%d%d%d", &opt, &u, &v);
        u ^= las, v ^= las;
        if(opt) printf("%d\n", las = query(u, v));
        else add(u, v, ++cnt);
    }
    return 0;
}

 

来源：https://www.cnblogs.com/lwqq3/p/11526912.html