【洛谷 P1501】 [国家集训队]Tree II(LCT)

题目链接

Tree Ⅱ(=)【模板】LCT+【模板】线段树2。。

分别维护3个标记,乘的时候要把加法标记也乘上。
还有就是模数的平方刚好爆(int),所以开昂赛德(int)就可以了。
我把初始化放在连边的那个循环里了,而那个循环是(1)(n-1)的,所以第(n)个没初始化到。。(WA)了好久。

#include <cstdio>
#include <cstring>
#define YCH 51061
#define R register unsigned int
#define I inline void
#define lc c[x][0]
#define rc c[x][1]
#define mul(x, y) x *= y; x %= YCH
#define add(x, y) x += y; x %= YCH
const int MAXN = 100010;
inline int read(){
    int s = 0, w = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9'){ if(ch == '-') w = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); }
    return s * w;
}
unsigned int f[MAXN], c[MAXN][2], v[MAXN], s[MAXN], st[MAXN], rt[MAXN], mt[MAXN], at[MAXN], sz[MAXN];
inline int nroot(R x){
	return c[f[x]][0] == x || c[f[x]][1] == x;
}
I pushup(R x){
	s[x] = (s[lc] + s[rc] + v[x]) % YCH;
	sz[x] = sz[lc] + sz[rc] + 1;
}
I pushr(R x){
	lc ^= rc; rc = lc ^ rc; lc ^= rc; rt[x] ^= 1;
}
I pushm(R x, R p){
	mul(s[x], p); mul(at[x], p);
	mul(mt[x], p); mul(v[x], p);
}
I pusha(R x, R p){
	add(s[x], p * sz[x]);
	add(at[x], p); add(v[x], p);
}
I pushdown(R x){
	if(mt[x] != 1){
		pushm(lc, mt[x]); pushm(rc, mt[x]);
		mt[x] = 1;
	}
	if(at[x]){
		pusha(lc, at[x]); pusha(rc, at[x]);
		at[x] = 0;
	}
	if(rt[x]){
		pushr(lc); pushr(rc);
		rt[x] = 0;
	}
}
I rotate(R x){
	R y = f[x], z = f[y], k = c[y][1] == x, w = c[x][!k];
	if(nroot(y)) c[z][c[z][1] == y] = x;
	c[x][!k] = y; c[y][k] = w; f[y] = x; f[x] = z;
	if(w) f[w] = y;
	pushup(y);
}
I splay(R x){
	R y = x, z = 0;
	st[++z] = y;
	while(nroot(y)) st[++z] = y = f[y];
	while(z) pushdown(st[z--]);
	while(nroot(x)){
		y = f[x]; z = f[y];
		if(nroot(y)) (c[z][1] == y) ^ (c[y][1] == x) ? rotate(x) : rotate(y);
		rotate(x);
	}
	pushup(x);
}
I access(R x){
	for(R y = 0; x; x = f[y = x]){
	   splay(x); rc = y; pushup(x);
    }
}
I makeroot(R x){
	access(x); splay(x); 
	pushr(x);
}
I split(R x, R y){
	makeroot(x); access(y); splay(y);
}
I link(R x, R y){
	makeroot(x);
	f[x] = y;
}
I cut(R x, R y){
	split(x, y);
	f[x] = c[y][0] = 0;
	pushup(y);
}
int n, m, a, b, x, y;
char opt;
int main(){
	n = read(); m = read();
	for(R i = 1; i < n; ++i)
	   link(read(), read()), v[i] = mt[i] = 1;
	v[n] = mt[n] = 1;
	while(m--){
		opt = getchar(); while(opt != '-' && opt != '+' && opt != '*' && opt != '/') opt = getchar();
		a = read(); b = read();
		switch(opt){
			case '+' : x = read(); split(a, b); pusha(b, x); break;
			case '-' : x = read(); y = read(); cut(a, b); link(x, y); break;
			case '*' : x = read(); split(a, b); pushm(b, x); break;
			case '/' : split(a, b); printf("%d
", s[b]); break;
		}
	}
	return 0;
}
原文地址:https://www.cnblogs.com/Qihoo360/p/10333424.html