bzoj3282: Tree

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ls c[x][0]
#define rs c[x][1]
using namespace std;
const int maxn=300010;
int n,m;
 
struct LCT{
    int c[maxn][2],fa[maxn],rev[maxn],sum[maxn],val[maxn];
    bool isroot(int x){return (c[fa[x]][0]!=x)&&(c[fa[x]][1]!=x);}
    int which(int x){return c[fa[x]][1]==x;}
    void update(int x){sum[x]=sum[c[x][0]]^sum[c[x][1]]^val[x];}
    void flip(int x){swap(c[x][0],c[x][1]),rev[x]^=1;}
    void down(int x){if (rev[x]) flip(c[x][0]),flip(c[x][1]),rev[x]=0;}
    void relax(int x){if (!isroot(x)) relax(fa[x]);down(x);}
    void rotate(int x){
        int y=fa[x],z=fa[y],nx=which(x),ny=which(y);
        fa[c[x][!nx]]=y,c[y][nx]=c[x][!nx];
        fa[x]=z;if (!isroot(y)) c[z][ny]=x;
        fa[y]=x,c[x][!nx]=y;update(y);
    }
    void splay(int x){
        relax(x);
        while (!isroot(x)){
            if (isroot(fa[x])) rotate(x);
            else if (which(x)==which(fa[x])) rotate(fa[x]),rotate(x);
            else rotate(x),rotate(x);
        }
        update(x);
    }
    void access(int x){for (int p=0;x;x=fa[x]) splay(x),fa[c[x][1]=p]=x,update(x),p=x;}
    void makeroot(int x){access(x),splay(x),flip(x);}
    void modify(int x,int v){val[x]=v,splay(x);}
    int findroot(int x){access(x),splay(x);for (;c[x][0];x=c[x][0]);return x;}
    void link(int a,int b){makeroot(a),fa[a]=b;}
    void cut(int a,int b){makeroot(a),access(b),splay(b);if (c[b][0]==a) c[b][0]=fa[a]=0;}
    int query(int a,int b){makeroot(a),access(b),splay(b);return sum[b];}
}T;
 
int main(){
    scanf("%d%d",&n,&m);
    for (int i=1,a;i<=n;i++) scanf("%d",&a),T.modify(i,a);
    for (int i=1,op,a,b;i<=m;i++){
        scanf("%d%d%d",&op,&a,&b);
        if (op==0) printf("%d\n",T.query(a,b));
        else if (op==1){if (!(T.findroot(a)==T.findroot(b))) T.link(a,b);}
        else if (op==2) T.cut(a,b);
        else T.modify(a,b);
    }
    return 0;
}

又是一道tree...

题目大意:就是说的直接的bzoj2819:NIM,加边删边,单点修改。

思路:LCT维护链上的异或和,注意题目中的不保证


posted @ 2015-06-22 21:39  orzpps  阅读(112)  评论(0编辑  收藏  举报