2631: tree

2631: tree

Time Limit: 30 Sec  Memory Limit: 128 MB
Submit: 3561  Solved: 1189
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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

Source

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一看就是LCT?标记的转移记得用矩阵乘法，，自己先推一下(WA惨了)然后这题丧心病狂卡常数啊。。。maintain可以少打一些(如rotate操作只需要一个maintain)最后，，本题请使用unsigned intlong long计算更慢些(日。。。)

#include<iostream>#include<cstdio>#include<queue>#include<vector>#include<bitset>#include<algorithm>#include<cstring>#include<map>#include<stack>#include<set>#include<cmath>#include<ext/pb_ds/priority_queue.hpp>using namespace std;const int maxn = 1E5 + 10;typedef unsigned int LL;const LL mo = 51061;int n,m,ch[maxn][2],rev[maxn],fa[maxn],pfa[maxn],vis[maxn];LL va[maxn],sum[maxn],Add[maxn],siz[maxn],Multi[maxn];char com[10];vector <int> v[maxn];queue <int> Q;void pushdown(int x){if (rev[x]) {swap(ch[x][0],ch[x][1]); rev[x] ^= 1;if (ch[x][0]) rev[ch[x][0]] ^= 1;if (ch[x][1]) rev[ch[x][1]] ^= 1;}if (Add[x] || Multi[x] != 1) {va[x] *= Multi[x]; va[x] %= mo;sum[x] *= Multi[x]; sum[x] %= mo;va[x] += Add[x]; va[x] %= mo;sum[x] += Add[x]*siz[x]; sum[x] %= mo;for (int i = 0; i < 2; i++)if (ch[x][i]) {Add[ch[x][i]] = Multi[x]*Add[ch[x][i]] + Add[x],Add[ch[x][i]] %= mo;Multi[ch[x][i]] *= Multi[x],Multi[ch[x][i]] %= mo;}Add[x] = 0; Multi[x] = 1;}}void maintain(int x){siz[x] = 1; sum[x] = va[x];for (int i = 0; i < 2; i++)if (ch[x][i]) {siz[x] += siz[ch[x][i]];sum[x] += sum[ch[x][i]]; sum[x] %= mo;}}void rotate(int x){int y = fa[x],z = fa[y];pfa[x] = pfa[y]; pfa[y] = 0;int d = ch[y][0] == x?0:1;ch[y][d] = ch[x][d^1];if (ch[x][d^1]) fa[ch[x][d^1]] = y; ch[x][d^1] = y; fa[y] = x; fa[x] = z;if (z) ch[z][ch[z][1] == y] = x;if (ch[y][0]) pushdown(ch[y][0]);if (ch[y][1]) pushdown(ch[y][1]);maintain(y);}int s[maxn],top;void splay(int x){top = 0;for (int now = x; now; now = fa[now]) s[++top] = now;while (top) {pushdown(s[top--]);}for (int y; y = fa[x]; rotate(x))if (fa[y])rotate((ch[fa[y]][0] == y)^(ch[y][0] == x)?x:y);if (ch[x][0]) pushdown(ch[x][0]);if (ch[x][1]) pushdown(ch[x][1]);maintain(x);}void Access(int x){for (int u = x,v = 0; u; v = u,u = pfa[u]) {splay(u);if (ch[u][1]) fa[ch[u][1]] = 0,pfa[ch[u][1]] = u;ch[u][1] = v; if (v) pfa[v] = 0,fa[v] = u;}}void ChangeRoot(int x) {Access(x); splay(x); rev[x] ^= 1;}void Join(int x,int y) {ChangeRoot(x); pfa[x] = y; splay(y);}void Cut(int x,int y) {ChangeRoot(x); Access(y); splay(y); ch[y][0] = fa[x] = 0;}int Query(int x,int y) {ChangeRoot(x); Access(y); splay(y); return sum[y];}void add(int x,int y,int c){ChangeRoot(x); Access(y); splay(y);Add[y] += 1LL*c; Add[y] %= mo; }void multi(int x,int y,int c){ChangeRoot(x); Access(y); splay(y);//Add[y] *= 1LL*c; Add[y] %= mo;Multi[y] *= 1LL*c; Multi[y] %= mo; }int getint(){char ch = getchar();int ret = 0;while (ch < '0' ||  '9' < ch) ch = getchar();while ('0' <= ch && ch <= '9') ret = ret*10 + ch - '0',ch = getchar();return ret;}char c[20];void print( int k )    {    if (!k) {puts("0"); return;}    int num = 0;        while( k > 0 ) c[++num] = k % 10, k /= 10;        while( num )             putchar( c[num--]+48 );        putchar( 10 );    }    int main(){freopen("a.in","r",stdin);freopen("a.out","w",stdout); n = getint(); m = getint();for (int i = 1; i <= n; i++) sum[i] = siz[i] = Multi[i] = va[i] = 1;for (int i = 1; i < n; i++) Join(getint(),getint());while (m--) {scanf("%s",com);int x = getint(),y = getint();if (com[0] == '+') {int c = getint(); add(x,y,c);}else if (com[0] == '-') {Cut(x,y); x = getint(); y = getint();Join(x,y);}else if (com[0] == '*') {int c = getint(); multi(x,y,c);}else print(Query(x,y));}return 0;}