[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 <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