BZOJ 4668 冷战

Posted 2020-08-05 ziliuziliu

这题首先应该想lct的做法：每次link一个递增的边权，查询两点边权最大值。

后来发现按秩合并的并查集可以搞。复杂度nlogn可过。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 500500
using namespace std;
int n,m,type,x,y,lastans=0,fath[maxn],rank[maxn],dep[maxn],times=0,val[maxn];
int getfather(int x)
{
    if (fath[x]==x) return x;
    int ret=getfather(fath[x]);
    dep[x]=dep[fath[x]]+1;
    return ret;
}
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=fath[x];
    }
    return ans;
}
int main()
{
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++)
        fath[i]=i;
    for (int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&type,&x,&y);
        x^=lastans;y^=lastans;
        if (type==0)
        {
            times++;
            int px=getfather(x),py=getfather(y);
            if (rank[px]>=rank[py])
            {
                fath[py]=px;
                val[py]=times;
                if (rank[px]==rank[py]) rank[px]++;
            }
            else
            {
                fath[px]=py;
                val[px]=times;
            }
        }
        else
        {
            int px=getfather(x),py=getfather(y);
            if (px!=py) {lastans=0;printf("%d\n",lastans);}
            else
            {
                lastans=lca(x,y);
                printf("%d\n",lastans);
            }
        }
    }
    return 0;
}