[COGS 1799][国家集训队2012]tree(伍一鸣)

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

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

题解

比较简单，用来练习 $lct$ 上的 $lazy$ 操作。 

  1 //It is made by Awson on 2018.1.16
  2 #include <set>
  3 #include <map>
  4 #include <cmath>
  5 #include <ctime>
  6 #include <queue>
  7 #include <stack>
  8 #include <cstdio>
  9 #include <string>
 10 #include <vector>
 11 #include <cstdlib>
 12 #include <cstring>
 13 #include <iostream>
 14 #include <algorithm>
 15 #define LL long long
 16 #define Abs(a) ((a) < 0 ? (-(a)) : (a))
 17 #define Max(a, b) ((a) > (b) ? (a) : (b))
 18 #define Min(a, b) ((a) < (b) ? (a) : (b))
 19 #define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
 20 using namespace std;
 21 const int MOD = 51061;
 22 const int N = 1e5;
 23 void read(int &x) {
 24     char ch; bool flag = 0;
 25     for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
 26     for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
 27     x *= 1-2*flag;
 28 }
 29 void write(int x) {
 30     if (x > 9) write(x/10);
 31     putchar(x%10+48);
 32 }
 33 
 34 char ch[10];
 35 int n, q, u, v, c;
 36 struct Link_Cut_Tree {
 37     int ch[N+5][2], pre[N+5], rev[N+5], sum[N+5], prod[N+5], val[N+5], tol[N+5], isrt[N+5], size[N+5];
 38     Link_Cut_Tree() {for (int i = 1; i <= N; i++) val[i] = tol[i] = isrt[i] = prod[i] = size[i] = 1; }
 39     void pushup(int o) {tol[o] = (tol[ch[o][0]]+tol[ch[o][1]]+val[o])%MOD, size[o] = (size[ch[o][0]]+size[ch[o][1]]+1)%MOD; }
 40     void pushdown(int o) {
 41     int ls = ch[o][0], rs = ch[o][1];
 42     if (rev[o]) {
 43         Swap(ch[ls][0], ch[ls][1]), Swap(ch[rs][0], ch[rs][1]);
 44         rev[ls] ^= 1, rev[rs] ^= 1, rev[o] = 0;
 45     }
 46     if (prod[o] != 1) {
 47         prod[ls] = (LL)prod[ls]*prod[o]%MOD, prod[rs] = (LL)prod[rs]*prod[o]%MOD;
 48         sum[ls] = (LL)sum[ls]*prod[o]%MOD, sum[rs] = (LL)sum[rs]*prod[o]%MOD;
 49         val[ls] = (LL)val[ls]*prod[o]%MOD, val[rs] = (LL)val[rs]*prod[o]%MOD;
 50         tol[ls] = (LL)tol[ls]*prod[o]%MOD, tol[rs] = (LL)tol[rs]*prod[o]%MOD;
 51         prod[o] = 1;
 52     }
 53     if (sum[o]) {
 54         sum[ls] = (sum[ls]+sum[o])%MOD, sum[rs] = (sum[rs]+sum[o])%MOD;
 55         val[ls] = (val[ls]+sum[o])%MOD, val[rs] = (val[rs]+sum[o])%MOD;
 56         tol[ls] = (tol[ls]+(LL)sum[o]*size[ls]%MOD)%MOD, tol[rs] = (tol[rs]+(LL)sum[o]*size[rs]%MOD)%MOD;
 57         sum[o] = 0;
 58     }
 59     }
 60     void push(int o) {
 61     if (!isrt[o]) push(pre[o]);
 62     pushdown(o);
 63     }
 64     void rotate(int o, int kind) {
 65     int p = pre[o];
 66     ch[p][!kind] = ch[o][kind], pre[ch[o][kind]] = p;
 67     if (isrt[p]) isrt[o] = 1, isrt[p] = 0;
 68     else ch[pre[p]][ch[pre[p]][1] == p] = o;
 69     pre[o] = pre[p];
 70     ch[o][kind] = p, pre[p] = o;
 71     pushup(p), pushup(o);
 72     }
 73     void splay(int o) {
 74     push(o);
 75     while (!isrt[o]) {
 76         if (isrt[pre[o]]) rotate(o, ch[pre[o]][0] == o);
 77         else {
 78         int p = pre[o], kind = ch[pre[p]][0] == p;
 79         if (ch[p][kind] == o) rotate(o, !kind), rotate(o, kind);
 80         else rotate(p, kind), rotate(o, kind);
 81         }
 82     }
 83     }
 84     void access(int o) {
 85     int y = 0;
 86     while (o) {
 87         splay(o);
 88         isrt[ch[o][1]] = 1, isrt[ch[o][1] = y] = 0;
 89         pushup(o); o = pre[y = o];
 90     }
 91     }
 92     void makeroot(int o) {access(o), splay(o); rev[o] ^= 1, Swap(ch[o][0], ch[o][1]); }
 93     void link(int x, int y) {makeroot(x); pre[x] = y; }
 94     void cut(int x, int y) {makeroot(x), access(y), splay(y); ch[y][0] = pre[x] = 0, isrt[x] = 1; pushup(y); }
 95     void split(int x, int y) {makeroot(x), access(y), splay(y); }
 96     void add(int x, int y, int c) {split(x, y); sum[y] = (sum[y]+c)%MOD, val[y] = (val[y]+c)%MOD, tol[y] = (tol[y]+(LL)c*size[y]%MOD)%MOD; }
 97     void plus(int x, int y, int c) {split(x, y); prod[y] = (LL)prod[y]*c%MOD, sum[y] = (LL)sum[y]*c%MOD, val[y] = (LL)val[y]*c%MOD, tol[y] = (LL)tol[y]*c%MOD; }
 98     int query(int x, int y) {split(x, y); return tol[y]; }
 99 }T;
100 
101 void work() {
102     read(n), read(q);
103     for (int i = 1; i < n; i++) {
104     read(u), read(v); T.link(u, v);
105     }
106     while (q--) {
107     scanf("%s", ch);
108     if (ch[0] == '+') {read(u), read(v), read(c); T.add(u, v, c); }
109     else if (ch[0] == '-') {
110         read(u), read(v); T.cut(u, v);
111         read(u), read(v); T.link(u, v);
112     }
113     else if (ch[0] == '*') {read(u), read(v), read(c); T.plus(u, v, c); }
114     else if (ch[0] == '/') {read(u), read(v); write(T.query(u, v)), putchar('\n'); }
115     }
116 }
117 int main() {
118     work();
119     return 0;
120 }

 

posted @ 2018-01-16 16:06  NaVi_Awson  阅读(207)  评论(0)  编辑  收藏  举报