[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 <bits/stdc++.h> 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