• BZOJ3510 首都(LCT)


      即动态维护树的重心。考虑合并后的新重心一定在两棵树的重心的连线上。于是对每个点维护其子树大小,合并时在这条链的splay上二分即可。至于如何维护子树大小,见https://blog.csdn.net/neither_nor/article/details/52979425。明明都看那么多题解说要注意pushdown,结果还是因为这个调了一年,心态爆炸。(下面代码复杂度可能是假的,因为dfs完没有splay叶子,反正也没人卡我就不管了)

    #include<iostream> 
    #include<cstdio>
    #include<cmath>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    #define ll long long
    #define N 100010
    #define M 200010
    #define lson tree[k].ch[0]
    #define rson tree[k].ch[1]
    #define lself tree[tree[k].fa].ch[0]
    #define rself tree[tree[k].fa].ch[1]
    char getc(){char c=getchar();while (c<'A'||c>'Z') c=getchar();return c;}
    int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
    int read()
    {
        int x=0,f=1;char c=getchar();
        while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
        while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
        return x*f;
    }
    int n,m,ans;
    struct data{int ch[2],fa,rev,size_f,size;
    }tree[N];
    void up(int k){tree[k].size=tree[lson].size+tree[rson].size+tree[k].size_f+1;}
    void rev(int k){if (k) swap(lson,rson),tree[k].rev^=1;}
    void down(int k){if (tree[k].rev) rev(lson),rev(rson),tree[k].rev=0;}
    bool isroot(int k){return lself!=k&&rself!=k;}
    int whichson(int k){return rself==k;}
    void push(int k){if (!isroot(k)) push(tree[k].fa);down(k);}
    void move(int k)
    {
        int fa=tree[k].fa,gf=tree[fa].fa,p=whichson(k);
        if (!isroot(fa)) tree[gf].ch[whichson(fa)]=k;tree[k].fa=gf;
        tree[tree[k].ch[!p]].fa=fa,tree[fa].ch[p]=tree[k].ch[!p];
        tree[k].ch[!p]=fa,tree[fa].fa=k;
        up(fa),up(k);
    }
    void splay(int k)
    {
        push(k);
        while (!isroot(k))
        {
            int fa=tree[k].fa;
            if (!isroot(fa))
                if (whichson(k)^whichson(fa)) move(k);
                else move(fa);
            move(k);
        }
    }
    void access(int k)
    {
        for (int t=0;k;t=k,k=tree[k].fa)
        {
            splay(k);
            tree[k].size_f+=tree[rson].size;
            rson=t;
            tree[k].size_f-=tree[rson].size;
            up(k);
        }
    }
    void makeroot(int k){access(k),splay(k),rev(k);}
    int findroot(int k){access(k),splay(k);for (;lson;k=lson) down(k);splay(k);return k;}
    int dfs(int k,int s,int r)
    {
        if (!k) return n+1;
        down(k);
        if (s-(tree[k].size-tree[lson].size+r)<=(s>>1))
        {
            int x=dfs(rson,s,r);
            if (x<=n) return x;else return k;
        }
        else return dfs(lson,s,r+tree[k].size-tree[lson].size);
    }
    int find(int k){down(k);if (rson) return find(rson);else return k;}
    void link(int x,int y)
    {
        int p=findroot(x),q=findroot(y);ans^=p,ans^=q;
        makeroot(y),access(x),splay(x),tree[y].fa=x;
        tree[x].size_f+=tree[y].size,up(x);
        makeroot(p),access(q),splay(q);int s=tree[q].size;
        p=dfs(q,s,0),splay(p);
        if (tree[p].size-tree[tree[p].ch[0]].size<=(s>>1))
        splay(q=find(tree[p].ch[0])),p=min(p,q);
        makeroot(p);ans^=p;
    }
    int main()
    {
    #ifndef ONLINE_JUDGE
        freopen("bzoj3510.in","r",stdin);
        freopen("bzoj3510.out","w",stdout);
        const char LL[]="%I64d
    ";
    #else
        const char LL[]="%lld
    ";
    #endif
        n=read(),m=read();
        for (int i=1;i<=n;i++) ans^=i,tree[i].size=1;
        while (m--)
        {
            char c=getc();
            if (c=='X') printf("%d
    ",ans);
            if (c=='A') link(read(),read());
            if (c=='Q') printf("%d
    ",findroot(read()));
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Gloid/p/10275008.html
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