牛客练习赛88 D 克鲁斯卡尔重构树

首先题目要求我们找的是 最小生成树上路径的最大值 倘若我们用普通的树上 lca去寻找的话 是需要logn级别去查询的 但是询问有1e7 所以我们考虑别的方法 思考到克鲁斯卡尔重构树 边权值小的在最下面 边权值大的在最上面 所以我们 可以使用lca o1查询 直接查询到 最大边权

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#define x first
#define y second

using namespace std;

const int N = 2e6 + 10,M = N * 2;

typedef unsigned long long ull;
ull myRand(ull &k1, ull &k2){
    
    ull k3 = k1, k4 = k2;
    k1 = k4;
    k3 ^= (k3 <<23);
    k2 = k3 ^ k4 ^ (k3 >>17) ^ (k4 >>26);
    return k2 + k4;
}
pair<int,int>myRanq(ull&k1,ull&k2,int MAXN){
    
    int x=myRand(k1,k2)%MAXN+1,y=myRand(k1,k2)%MAXN+1;
    if(x>y)return make_pair(y,x);
    else return make_pair(x,y);
}

int head[N],to[M],last[M],cnt;
void add(int a,int b){
    
    to[++cnt] = b;
    last[cnt] = head[a];
    head[a] = cnt;
}

int fa[N];
int get(int x){
    
    if(fa[x] == x) return x;
    return fa[x] = get(fa[x]);
}

struct node{
    
    int x,y,w;
    bool operator < (const node &p)const{
    
       return w < p.w;
    }
}edge[N];

int depth[N];
int MIN(int x,int y){
    
    if(depth[x] < depth[y]) return x;
    return y;
}

int n,m,q,ext,v[N];
int st[N][20];
int eu[N],ans,id[N];
void dfs(int x,int pre){
    
    depth[x] = depth[pre] + 1;
    eu[++ans] = x,id[x] = ans;
    for(int i = head[x]; i != -1; i = last[i]){
    
        int j = to[i];
        dfs(j,x);
        eu[++ans] = x;
    }
}

int LOG[N];
int lca(int x,int y){
    
    if(x > y) swap(x,y);
    int k = LOG[y - x + 1];
    return MIN(st[x][k],st[y - (1 << k) + 1][k]);
}

int main(){
    
    memset(head,-1,sizeof head);
    cin >> n >> m;
    for(int i = 1; i <= m; i++){
    
       scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].w);
    }
    sort(edge + 1,edge + m + 1);

    for(int i = 1; i <= n + m; i++) fa[i] = i;

    ext = n;
    for(int i = 1; i <= m; i++){
    
        int x = get(edge[i].x),y = get(edge[i].y);
        if(x != y){
    
            fa[x] = fa[y] = ++ext;
            v[ext] = edge[i].w;
            add(ext,x);
            add(ext,y);
        }
    }

    dfs(get(1),0);

    for(int i = 1; i <= ans; i++) st[i][0] = eu[i];
    for(int i = 2; i <= ans; i++) LOG[i] = LOG[i >> 1] + 1;
    for(int j = 1; j <= 19; j++){
    
        for(int i = 1; i + (1 << j) - 1 <= ans; i++){
    
            st[i][j] = MIN(st[i][j - 1],st[i + (1 << (j - 1))][j - 1]);
        }
    }
    int ANS = 0;
    int q;ull a1,a2;
    cin >> q >> a1 >> a2;
    while(q--){
    
        pair<int,int> p;
        p = myRanq(a1,a2,n);
        int x = p.x,y = p.y;
        ANS ^= v[lca(id[x],id[y])];

    }

    cout <<ANS <<endl;








    return 0;
}