[BZOJ2049][[Sdoi2008]Cave 洞穴勘测][LCT]

[BZOJ2049][[Sdoi2008]Cave 洞穴勘测][LCT]

思路：

题目大意就不放了，貌似是一道LCT裸题。。。三个操作分别是 link ， cut 和 find 。

感谢这篇集训队论文:https://wenku.baidu.com/view/75906f160b4e767f5acfcedb

以及黄学长的代码：http://hzwer.com/3921.html

感觉LCT的复杂度全靠 splay 和 access 的两个均摊啊。。。

以及由于 splay 的维护是按深度为序，所以把一个点提到根后它的父亲肯定是在它的左儿子中，所以只要一直向左找儿子就好了。

代码：

#include 
using namespace std;
const int Maxn = 10005;
namespace IO {
    inline char get(void) {
        static char buf[1000000], *p1 = buf, *p2 = buf;
        if (p1 == p2) {
            p2 = (p1 = buf) + fread(buf, 1, 1000000, stdin);
            if (p1 == p2) return EOF;
        }
        return *p1++;
    }
    inline void read(int &x) {
        x = 0; static char c; bool f = 0;
        for (; !(c >= '0' && c <= '9'); c = get()) if (c == '-') f = 1;
        for (; c >= '0' && c <= '9'; x = x * 10 + c - '0', c = get()); if (f) x = -x;
    }
    inline void read(char &x) {
        x = get();
        while(!(x>= 'A' && x <= 'Z')) x = get(); 
    }
    inline void write(int x) {
        if (!x) return (void)puts("0");
        if (x < 0) putchar('-'), x = -x;
        static short s[12], t;
        while (x) s[++t] = x % 10, x /= 10;
        while (t) putchar('0' + s[t--]);
        putchar('\n');
    }
};

int fa[Maxn], c[Maxn][2], st[Maxn], top;
bool rev[Maxn];
inline bool isRt(int x) {
    return c[fa[x]][0] != x && c[fa[x]][1] != x;
}
inline void pushDown(int x) {
    static int l, r;
    l = c[x][0], r = c[x][1];
    if (rev[x]) {
        rev[l] ^= 1; rev[r] ^= 1;
        swap(c[x][0], c[x][1]);
        rev[x] ^= 1;
    }
}
inline void rotate(int x) {
    int y = fa[x], z = fa[y], l, r;
    r = c[y][0] == x; l = r ^ 1;
    if (!isRt(y)) c[z][c[z][1] == y] = x;
    fa[x] = z; fa[y] = x; fa[c[x][r]] = y;
    c[y][l] = c[x][r]; c[x][r] = y;
}
inline void splay(int x) {
    st[top = 1] = x;
    for (int i = x; !isRt(i); i = fa[i]) st[++top] = fa[i];
    while (top) pushDown(st[top--]);
    int y, z;
    while (!isRt(x)) {
        y = fa[x], z = fa[y];
        if (!isRt(y)) c[y][0] == x ^ c[z][0] == y ? rotate(x) : rotate(y);
        rotate(x);
    }
}
inline void access(int x) {
    int t = 0;
    while (x) {
        splay(x);
        c[x][1] = t;
        t = x; x = fa[x];
    }
}
inline void rever(int x) {
    access(x); splay(x); rev[x] ^= 1;
}
inline void link(int x, int y) {
    rever(x); fa[x] = y; splay(x);
}
inline void cut(int x, int y) {
    rever(x); access(y); splay(y); c[y][0] = fa[x] = 0;
}
inline int find(int x) {
    access(x); splay(x);
    int y = x;
    while (c[y][0]) y = c[y][0];
    return y;
}
int n, m;
int main(void) {
    //freopen("in.txt", "r", stdin);
    IO::read(n), IO::read(m);
    char op; int x, y;
    while (m--) {
        IO::read(op), IO::read(x), IO::read(y);
        if (op == 'C') link(x, y);
        else if (op == 'D') cut(x, y);
        else puts(find(x) == find(y) ? "Yes" : "No");
    }
    return 0;
}

完。

By g1n0st