BZOJ 2049 洞穴勘测

LCT判断联通性
没什么特别的。。还是一个普通的板子题,把LCT当并查集用了,只不过LCT灵活一些,还可以断边

话说自从昨天被维修数列那题榨干之后我现在写splay都不用动脑子了,,机械式的码splay23333

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 10005;
int n, m, tot, ch[N][2], fa[N], rev[N], st[N];

int newNode(){
    fa[++tot] = 0; ch[tot][0] = ch[tot][1] = 0;
    return tot;
}

bool isRoot(int x){
    return ch[fa[x]][0] != x && ch[fa[x]][1] != x;
}

void reverse(int x){
    rev[x] ^= 1;
    swap(ch[x][0], ch[x][1]);
}

void push_down(int x){
    if(rev[x]){
        rev[x] ^= 1;
        reverse(ch[x][0]), reverse(ch[x][1]);
    }
}

void rotate(int x){
    int y = fa[x], z = fa[y], p = (ch[y][1] == x) ^ 1;
    push_down(y), push_down(x);
    ch[y][p^1] = ch[x][p], fa[ch[x][p]] = y;
    if(!isRoot(y)) ch[z][ch[z][1] == y] = x;
    fa[x] = z, fa[y] = x, ch[x][p] = y;
}

void splay(int x){
    int pos = 0; st[++pos] = x;
    for(int i = x; !isRoot(i); i = fa[i]) st[++pos] = fa[i];
    while(pos) push_down(st[pos--]);
    while(!isRoot(x)){
        int y = fa[x], z = fa[y];
        if(!isRoot(y)){
            if((ch[y][0] == x) ^ (ch[z][0] == y)) rotate(x);
            else rotate(y);
        }
        rotate(x);
    }
}

void access(int x){
    for(int p = 0; x; p = x, x = fa[x])
        splay(x), ch[x][1] = p;
}

void makeRoot(int x){
    access(x), splay(x), reverse(x);
}

int findRoot(int x){
    access(x), splay(x);
    while(ch[x][0]) x = ch[x][0];
    return x;
}

void link(int x, int y){
    makeRoot(x);
    if(findRoot(y) != x) fa[x] = y;
}

void cut(int x, int y){
    makeRoot(x), access(y), splay(y);
    fa[x] = ch[y][0] = 0;
}

bool isConnect(int x, int y){
    return findRoot(x) == findRoot(y);
}

int main(){

    int n = read(), m = read();
    for(int i = 1; i <= n; i ++) newNode();
    while(m --){
        char opt[20]; scanf("%s", opt);
        int x = read(), y = read();
        if(opt[0] == 'Q') printf(isConnect(x, y) ? "Yes\n" : "No\n");
        else if(opt[0] == 'C') link(x, y);
        else if(opt[0] == 'D') cut(x, y);
    }
    return 0;
}
posted @ 2019-04-13 16:38  清楚少女ひなこ  阅读(139)  评论(0编辑  收藏  举报