题意
一棵树,有删边加边,有一条链加/乘一个数,有询问一条链的和
分析
LCT,像线段树一样维护两个标记(再加上翻转标记就是三个),维护size,就行了
CODE
#include<bits/stdc++.h>
using namespace std;
inline void read(int &num) {
char ch; int flg = 1; while(!isdigit(ch=getchar()))if(ch=='-')flg = -flg;
for(num=0; isdigit(ch); num=num*10+ch-'0', ch=getchar()); num*=flg;
}
const int MAXN = 100005;
const int mod = 51061;
namespace LCT {
#define ls ch[x][0]
#define rs ch[x][1]
int ch[MAXN][2], fa[MAXN], w[MAXN], sz[MAXN], mul[MAXN], add[MAXN], sum[MAXN];
bool rev[MAXN];
inline bool isr(int x) { return ch[fa[x]][0] != x && ch[fa[x]][1] != x; }
inline bool get(int x) { return ch[fa[x]][1] == x; }
inline void upd(int x) {
sum[x] = (sum[ls] + sum[rs] + w[x]) % mod;
sz[x] = sz[ls] + sz[rs] + 1;
}
inline void pusha(int x, int val) {
if(!x) return;
sum[x] = (sum[x] + 1ll * val * sz[x]) % mod;
w[x] = (w[x] + val) % mod;
add[x] = (add[x] + val) % mod;
}
inline void pushm(int x, int val) {
if(!x) return;
sum[x] = 1ll * sum[x] * val % mod;
w[x] = 1ll * w[x] * val % mod;
add[x] = 1ll * add[x] * val % mod;
mul[x] = 1ll * mul[x] * val % mod;
}
inline void mt(int x) {
if(rev[x]) rev[ls]^=1, rev[rs]^=1, rev[x]^=1, swap(ls, rs);
if(mul[x]!=1) pushm(ls, mul[x]), pushm(rs, mul[x]), mul[x] = 1;
if(add[x]) pusha(ls, add[x]), pusha(rs, add[x]), add[x] = 0;
}
void mtpath(int x) { if(!isr(x)) mtpath(fa[x]); mt(x); }
inline void rot(int x) {
int y = fa[x], z = fa[y]; bool l = get(x), r = l^1;
if(!isr(y)) ch[z][get(y)] = x;
fa[ch[x][r]] = y; fa[y] = x; fa[x] = z;
ch[y][l] = ch[x][r]; ch[x][r] = y;
upd(y), upd(x);
}
inline void splay(int x) {
mtpath(x);
for(; !isr(x); rot(x))
if(!isr(fa[x])) rot(get(fa[x])==get(x)?fa[x]:x);
}
inline int access(int x) { int y = 0;
for(; x; x=fa[y=x]) splay(x), ch[x][1] = y, upd(x);
return y;
}
inline void bert(int x) { access(x), splay(x), rev[x]^=1; }
inline int sert(int x) {
access(x), splay(x);
for(; ch[x][0]; x=ch[x][0]);
return x;
}
inline void Link(int x, int y) { bert(x); if(sert(y) != x) fa[x] = y; }
inline void Cut(int x, int y) { bert(x); if(sert(y) == x && fa[x] == y && !ch[x][1]) fa[x] = ch[y][0] = 0, upd(y); }
inline int split(int x, int y) { bert(x), access(y), splay(y); return y; }
inline void Add(int x, int y, int val) { pusha(split(x, y), val); }
inline void Mul(int x, int y, int val) { pushm(split(x, y), val); }
inline int Query(int x, int y) { return sum[split(x, y)]; }
#undef ls
#undef rs
}
using namespace LCT;
int n, q;
int main() {
//freopen("sample.txt", "r", stdin);
read(n), read(q);
for(int i = 1; i <= n; ++i)
w[i] = sum[i] = mul[i] = sz[i] = 1;
for(int i = 1, x, y; i < n; ++i)
read(x), read(y), Link(x, y);
char ch[2];
int x, y, u, v;
while(q--) {
scanf("%s", ch);
switch(ch[0]) {
case '+':
read(u), read(v), read(x), Add(u, v, x%mod);
break;
case '-':
read(u), read(v), read(x), read(y), Cut(u, v), Link(x, y);
break;
case '*':
read(u), read(v), read(x), Mul(u, v, x%mod);
break;
case '/':
read(u), read(v), printf("%d
", Query(u, v));
break;
}
}
}