[SDOI2008]洞穴勘测【LCT维护联通关系】

题目链接


  LCT判断两点联通的这样的一个基础问题,因为不存在环,所以直接LCT维护连接关系即可。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e4 + 7;
int N, M;
namespace LCT
{
    int fa[maxN], c[maxN][2];
    int r[maxN];
    bool isroot(int x) { return c[fa[x]][0] != x && c[fa[x]][1] != x; }
    void pushup(int x)
    {
        
    }
    void pushr(int x) { swap(c[x][0], c[x][1]); r[x] ^= 1; }
    void pushdown(int x)
    {
        if(r[x])
        {
            if(c[x][0]) pushr(c[x][0]);
            if(c[x][1]) pushr(c[x][1]);
            r[x] = 0;
        }
    }
    void Rotate(int x)
    {
        int y = fa[x], z = fa[y], k = c[y][1] == x;
        if(!isroot(y)) c[z][c[z][1] == y] = x;
        fa[x] = z;
        c[y][k] = c[x][k ^ 1];
        fa[c[x][k ^ 1]] = y;
        c[x][k ^ 1] = y;
        fa[y] = x;
        pushup(y);
        pushup(x);
        if(z) pushup(z);
    }
    int Stap[maxN];
    void Splay(int x)
    {
        int y = x, z = 0;
        Stap[++z] = y;
        while(!isroot(y)) Stap[++z] = y = fa[y];
        while(z) pushdown(Stap[z--]);
        while(!isroot(x))
        {
            y = fa[x]; z = fa[y];
            if(!isroot(y)) (c[z][0] == y) ^ (c[y][0] == x) ? Rotate(x) : Rotate(y);
            Rotate(x);
        }
    }
    void Access(int x)
    {
        int y = 0;
        while(x)
        {
            Splay(x);
            c[x][1] = y;
            pushup(x);
            y = x;
            x = fa[x];
        }
    }
    void makeroot(int x)
    {
        Access(x);
        Splay(x);
        pushr(x);
    }
    int findroot(int x)
    {
        Access(x);
        Splay(x);
        while(c[x][0])
        {
            pushdown(x);
            x = c[x][0];
        }
        Splay(x);
        return x;
    }
    void Split(int x, int y)
    {
        makeroot(x);
        Access(y);
        Splay(y);
    }
    void link(int x, int y)
    {
        makeroot(x);
        if(findroot(y) != x)
        {
            fa[x] = y;
        }
    }
    void cut(int x, int y)
    {
        makeroot(x);
        if(findroot(y) != x || fa[y] != x || c[y][0]) return;
        fa[y] = c[x][1] = 0;
        pushup(x);
    }
};
using namespace LCT;
int main()
{
    scanf("%d%d", &N, &M);
    char ch[10]; int u, v;
    while(M--)
    {
        scanf("%s%d%d", ch, &u, &v);
        switch (ch[0])
        {
            case 'Q':
            {
                if(findroot(u) == findroot(v)) printf("Yes\n");
                else printf("No\n");
                break;
            }
            case 'C':
            {
                link(u, v);
                break;
            }
            default:
            {
                cut(u, v);
                break;
            }
        }
    }
    return 0;
}

 

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