【bzoj2631】tree LCT
题目描述
一棵n个点的树,每个点的初始权值为1。对于这棵树有q个操作,每个操作为以下四种操作之一:
+ u v c:将u到v的路径上的点的权值都加上自然数c;
- u1 v1 u2 v2:将树中原有的边(u1,v1)删除,加入一条新边(u2,v2),保证操作完之后仍然是一棵树;
* u v c:将u到v的路径上的点的权值都乘上自然数c;
/ u v:询问u到v的路径上的点的权值和,求出答案对于51061的余数。
输入
第一行两个整数n,q
接下来n-1行每行两个正整数u,v,描述这棵树
接下来q行,每行描述一个操作
接下来n-1行每行两个正整数u,v,描述这棵树
接下来q行,每行描述一个操作
输出
对于每个/对应的答案输出一行
样例输入
3 2
1 2
2 3
* 1 3 4
/ 1 1
样例输出
4
题解
带点权的LCT
需要注意的是3标记的处理:先乘后加,同时乘标记,与翻转互不影响。
这题的坑点在于int会WA,long long会TLE,必须用unsigned int。
#include <cstdio> #include <algorithm> #define N 100010 #define MOD 51061 #define lson c[0][x] #define rson c[1][x] using namespace std; int fa[N] , c[2][N] , si[N] , rev[N]; unsigned w[N] , sum[N] , add[N] , mul[N]; char str[5]; inline int read() { int ret = 0; char ch = getchar(); while(ch < '0' || ch > '9') ch = getchar(); while(ch >= '0' && ch <= '9') ret = (ret << 3) + (ret << 1) + ch - '0' , ch = getchar(); return ret; } void pushup(int x) { si[x] = si[lson] + si[rson] + 1; sum[x] = (sum[lson] + sum[rson] + w[x]) % MOD; } void cal(int x , unsigned a , unsigned m , int r) { sum[x] = (sum[x] * m + si[x] * a) % MOD; w[x] = (w[x] * m + a) % MOD; mul[x] = (mul[x] * m) % MOD; add[x] = (add[x] * m + a) % MOD; if(r) swap(lson , rson) , rev[x] ^= 1; } void pushdown(int x) { cal(lson , add[x] , mul[x] , rev[x]); cal(rson , add[x] , mul[x] , rev[x]); add[x] = rev[x] = 0 , mul[x] = 1; } bool isroot(int x) { return c[0][fa[x]] != x && c[1][fa[x]] != x; } void update(int x) { if(!isroot(x)) update(fa[x]); pushdown(x); } void rotate(int x) { int y = fa[x] , z = fa[y] , l = (c[1][y] == x) , r = l ^ 1; if(!isroot(y)) c[c[1][z] == y][z] = x; fa[x] = z , fa[y] = x , fa[c[r][x]] = y , c[l][y] = c[r][x] , c[r][x] = y; pushup(y) , pushup(x); } void splay(int x) { update(x); while(!isroot(x)) { int y = fa[x] , z = fa[y]; if(!isroot(y)) { if((c[0][y] == x) ^ (c[0][z] == y)) rotate(x); else rotate(y); } rotate(x); } } void access(int x) { int t = 0; while(x) splay(x) , rson = t , pushup(x) , t = x , x = fa[x]; } void makeroot(int x) { access(x) , splay(x); swap(lson , rson) , rev[x] ^= 1; } void link(int x , int y) { makeroot(x) , fa[x] = y; } void cut(int x , int y) { makeroot(x) , access(y) , splay(y) , c[0][y] = fa[x] = 0 , pushup(y); } void split(int x , int y) { makeroot(y) , access(x) , splay(x); } int main() { int n , m , i , x , y; unsigned z; n = read() , m = read(); for(i = 1 ; i <= n ; i ++ ) si[i] = w[i] = sum[i] = mul[i] = 1; for(i = 1 ; i < n ; i ++ ) x = read() , y = read() , link(x , y); while(m -- ) { scanf("%s" , str) , x = read() , y = read(); switch(str[0]) { case '+': z = (unsigned)read() , split(x , y) , cal(x , z , 1 , 0); break; case '-': cut(x , y) , x = read() , y = read() , link(x , y); break; case '*': z = (unsigned)read() , split(x , y) , cal(x , 0 , z , 0); break; default: split(x , y) , printf("%u\n" , sum[x]); } } return 0; }