[BZOJ2631]tree

tree

Description
　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。
Input
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
Output
　　对于每个/对应的答案输出一行
Sample Input
3 2
1 2
2 3
* 1 3 4
/ 1 1
Sample Output
4
HINT
数据规模和约定
10%的数据保证，1<=n，q<=2000
另外15%的数据保证，1<=n，q<=5*10^4，没有-操作，并且初始树为一条链
另外35%的数据保证，1<=n，q<=5*10^4，没有-操作
100%的数据保证，1<=n，q<=10^5，0<=c<=10^4

Solution
注意处理好tag

Code

#include <bits/stdc++.h>using namespace std;#define rep(i, l, r) for (int i = (l); i <= (r); i++)typedef unsigned int ui;template<typename T> inline void read(T &x){    x = 0; T f = 1; char ch = getchar();    while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }    while (isdigit(ch))  { x = x * 10 + ch - '0'; ch = getchar(); }    x *= f;}const int MOD = 51061;const int N = 100100;struct Node{    ui sz, sum, add_tag, mul_tag, rev_tag, val;    Node *c[2], *fa;    inline int d(){        if (fa->c[0] != this && fa->c[1] != this) return -1;        return fa->c[1] == this;    }    inline void sc(Node *p, int d) { c[d] = p; p->fa = this; }    inline void push_up(){        sz = c[0]->sz + c[1]->sz + 1;        sum = (c[0]->sum + c[1]->sum + val) % MOD;    }    inline void rev(){ swap(c[0], c[1]); rev_tag ^= 1; }    inline void mul(ui dt){        (val *= dt) %= MOD; (sum *= dt) %= MOD;        (mul_tag *= dt) %= MOD; (add_tag *= dt) %= MOD;    }    inline void add(ui dt){        (val += dt) %= MOD; (sum += sz % MOD * dt) %= MOD;        (add_tag += dt) %= MOD;    }    void push_down();}pool[N<<1], *null=pool, *tail=pool+1, *loc[N];void Node:: push_down(){    if (this->d() != -1) fa->push_down();    if (rev_tag){        rep(i, 0, 1) if (c[i] != null) c[i]->rev();        rev_tag ^= 1;    }    if (mul_tag != 1){        rep(i, 0, 1) if (c[i] != null) c[i]->mul(mul_tag);        mul_tag = 1;    }    if (add_tag){        rep(i, 0, 1) if (c[i] != null) c[i]->add(add_tag);        add_tag = 0;    }}int n, q;inline void setnull(){    null->c[0] = null->c[1] = null->fa = null;    null->mul_tag = 1;    null->rev_tag = null->add_tag = null->sum = null->sz = null->val = 0;}inline Node *newNode(ui w){    tail->c[0] = tail->c[1] = tail->fa = null;    tail->mul_tag = w;    tail->rev_tag = tail->add_tag = tail->sum = w; tail->sz = tail->val = 1;    return tail++;}void rot(Node *x){    Node *y = x->fa; int d = x->d();    if (y->d() == -1) x->fa = y->fa; else y->fa->sc(x, y->d());    y->sc(x->c[!d], d); y->push_up();    x->sc(y, !d);}void splay(Node *x){    for (x->push_down(); x->d() != -1; ){        if (x->fa->d() == -1) rot(x);        else x->d() == x->fa->d() ?             (rot(x->fa), rot(x)) : (rot(x), rot(x));    }     x->push_up();}void access(Node *x){    for (Node *t = null; x != null; t = x, x = x->fa)        splay(x), x->c[1] = t, x->push_up();}void mkroot(Node *x){access(x); splay(x); x->rev();}void link(Node *x, Node *y){mkroot(x); x->fa = y;}void cut(Node *x, Node *y){mkroot(x); access(y); splay(y); y->c[0]=x->fa=null; y->push_up();}void split(Node *x, Node *y){mkroot(x); access(y); splay(y);}int main(){    setnull();    read(n); read(q);    for (int i = 1; i <= n; i++) loc[i] = newNode(1);    for (int i = 1; i < n; i++){        int u, v; read(u); read(v); link(loc[u], loc[v]);    }       while (q--){        char opt; scanf(" %c", &opt);        int u, v, x, y; ui c; read(u); read(v);        if (opt == '+'){ read(c); split(loc[u], loc[v]); loc[v]->add(c); }        else if (opt == '-'){ read(x); read(y); cut(loc[u], loc[v]); link(loc[x], loc[y]); }        else if (opt == '*'){ read(c); split(loc[u], loc[v]); loc[v]->mul(c); }        else if (opt == '/'){ split(loc[u], loc[v]); printf("%u\n", loc[v]->sum);}    }    return 0;}