BZOJ4668——冷战

1、题意：给一些点，有两种操作， 连通两个点和询问两个点最早是在什么时候连通
2、分析：这题用lct维护动态最小生成树就好。。但是过于暴力，我们可以倒着用并查集暴力查询，然后依旧是查询链上的最大值，然而这次我们不路径压缩，我们按秩合并，然后树高$logn$，接下来就随意弄啦

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define M 1000010

inline int read(){
    char ch = getchar(); int x = 0, f = 1;
    while(ch < ‘0‘ || ch > ‘9‘){
        if(ch == ‘-‘) f = -1;
        ch = getchar();
    }
    while(‘0‘ <= ch && ch <= ‘9‘){
        x = x * 10 + ch - ‘0‘;
        ch = getchar();
    }
    return x * f;
}

int fa[M], Rank[M], dep[M], val[M]; 

inline int find(int x){
    if(fa[x] == x) return x;
    int ret = find(fa[x]);
    dep[x] = dep[fa[x]] + 1;
    return ret;
}

inline int lca(int x, int y){
    int ans = 0;
    while(x != y){
        if(dep[x] < dep[y]) swap(x, y);
        ans = max(ans, val[x]);
        x = fa[x];
    }
    return ans;
}

int main(){
    //freopen("0input.in", "r", stdin); 
    int n = read(), m = read();
    for(int i = 1; i <= n; i ++){
        fa[i] = i; Rank[i] = 1; dep[i] = 0;
    }
    int time = 0, ans = 0;
    for(int i = 1; i <= m; i ++){
        int op = read(), x = read(), y = read();
        x ^= ans; y ^= ans;
        if(!op){
            int px = find(x), py = find(y);
            time ++;
            if(px != py){
                if(Rank[px] >= Rank[py]){
                    fa[py] = px;
                    val[py] = time;
                    if(Rank[px] == Rank[py]) Rank[px] ++;
                }
                else{
                    fa[px] = py;
                    val[px] = time;
                }
            }
        }
        else{
            int px = find(x), py = find(y);
            if(px != py) ans = 0;
            else ans = lca(x, y);
            printf("%d\n", ans);
        }
    }
    return 0;
}
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