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。
输入格式:
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
输出格式:
对于每个/对应的答案输出一行
输入样例#1: 复制
3 2
1 2
2 3
* 1 3 4
/ 1 1
输出样例#1: 复制
4
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
By (伍一鸣)
全程自己YY,调了一下午真累啊QWQ......
LCT的板子题,要维护子树和,加法标记,乘法标记和自身的值
放标记的时候先放乘法标记
至于为什么http://www.cnblogs.com/zwfymqz/p/8588693.html
// luogu-judger-enable-o2
#include<iostream>
#include<cstdio>
#include<cstring>
#define int long long
const int mod = 51061;
using namespace std;
const int MAXN = 1e5+10;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
return x * f;
}
int N, Q;
#define ls(x) T[x].ch[0]
#define rs(x) T[x].ch[1]
#define fa(x) T[x].f
struct node {
int f, ch[2], r, mul, add, siz, sum, val;
}T[MAXN];
bool IsRoot(int x) {
return T[fa(x)].ch[0] != x && T[fa(x)].ch[1] != x;
}
int ident(int x) {
return T[fa(x)].ch[0] == x ? 0 : 1;
}
void connect(int x, int fa, int how) {
T[x].f = fa;
T[fa].ch[how] = x;
}
void update(int x) {
T[x].sum = (T[ls(x)].sum + T[rs(x)].sum + T[x].val ) % mod;
T[x].siz = T[ls(x)].siz + T[rs(x)].siz + 1 ;
}
void rotate(int x) {
int Y = fa(x), R = fa(Y), Yson = ident(x), Rson = ident(Y);
int B = T[x].ch[Yson ^ 1];
T[x].f = R;
if(!IsRoot(Y))
connect(x, R, Rson);
connect(B, Y, Yson);
connect(Y, x, Yson ^ 1);
update(Y);update(x);
}
void pushr(int x) {
if(T[x].r) {
swap(ls(x), rs(x));
T[ls(x)].r ^= 1;
T[rs(x)].r ^= 1;
T[x].r = 0;
}
}
void pushmul(int x) {
T[ls(x)].val *= T[x].mul; T[ls(x)].val %= mod;
T[rs(x)].val *= T[x].mul; T[rs(x)].val %= mod;
T[ls(x)].sum *= T[x].mul; T[ls(x)].sum %= mod;
T[rs(x)].sum *= T[x].mul; T[rs(x)].sum %= mod;
T[ls(x)].add *= T[x].mul; T[ls(x)].add %= mod;
T[rs(x)].add *= T[x].mul; T[rs(x)].add %= mod;
T[ls(x)].mul *= T[x].mul; T[ls(x)].mul %= mod;
T[rs(x)].mul *= T[x].mul; T[rs(x)].mul %= mod;
T[x].mul = 1;
}
void pushadd(int x) {
T[ls(x)].val += T[x].add; T[ls(x)].val %= mod;
T[rs(x)].val += T[x].add; T[rs(x)].val %= mod;
T[ls(x)].sum += T[ls(x)].siz * T[x].add; T[ls(x)].sum %= mod;
T[rs(x)].sum += T[rs(x)].siz * T[x].add; T[rs(x)].sum %= mod;
T[ls(x)].add += T[x].add; T[ls(x)].add %= mod;
T[rs(x)].add += T[x].add; T[rs(x)].add %= mod;
T[x].add = 0;
}
void pushdown(int x) {
pushr(x);
pushmul(x);
pushadd(x);
}
int st[MAXN];
int fuck;
void splay(int x) {
int y = x, top = 0;
st[++top] = y;
while(!IsRoot(y)) st[++top] = (y = fa(y));
while(top) pushdown(st[top--]);
for(int y = fa(x); !IsRoot(x); rotate(x), y = fa(x))
if(!IsRoot(y))
rotate( ident(x) == ident(y) ? y : x );
}
void access(int x) {
for(int y = 0; x; x = fa(y = x))//tag
splay(x), rs(x) = y, update(x);
}
void makeroot(int x) {
access(x);
splay(x);
T[x].r ^= 1;
}
void split(int x,int y) {
makeroot(x);
access(y);
splay(y);
}
void link(int x, int y) {
makeroot(x);
T[x].f = y;
}
int findroot(int x) {
access(x);
splay(x);
while(ls(x)) x = ls(x);
return x;
}
void cut(int x, int y) {
split(x, y);
//makeroot(x);
if(findroot(y) == x && fa(x) == y && !rs(x))
T[x].f = T[y].ch[0] = 0, update(y);
}
main() {
freopen("a.in","r",stdin);
scanf("%lld%lld", &N, &Q);
for(int i = 1; i <= N; i++)
T[i].val = 1, T[i].sum = 1, T[i].mul = 1, T[i].siz = 1, T[i].add = 0;
for(int i = 1; i <= N - 1; i++) {
int x = read(), y = read();
link(x, y);
}
int fuck = 0;
while(Q--) {
char c = 0; int u1, v1, u2, v2, val;
while(c < '*') c = getchar();
scanf("%lld%lld", &u1, &v1);
if(c == '+') {
scanf("%lld", &val);
split(u1, v1);
T[v1].val += val; T[v1].val %= mod;
T[v1].add += val; T[v1].add %= mod;
T[v1].sum += T[v1].siz * val; T[v1].sum %= mod;
}
else if(c == '-') {
scanf("%lld%lld", &u2, &v2);
cut(u1, v1);
link(u2, v2);
}
else if(c == '/') {
split(u1, v1);
printf("%d\n",T[v1].sum%mod);
}
else {
scanf("%lld",&val);
split(u1, v1);
T[v1].val *= val; T[v1].val %= mod;
T[v1].sum *= val; T[v1].siz %= mod;
T[v1].add *= val; T[v1].add %= mod;
T[v1].mul *= val; T[v1].mul %= mod;
}
}
return 0;
}