# Little Devil I

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 894    Accepted Submission(s): 296

Problem Description
There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil. Also, this devil is looking like a very cute Loli.

The devil likes to make thing in chaos. This kingdom’s road system is like simply a tree(connected graph without cycle). A road has a color of black or white. The devil often wants to make some change of this system.

In details, we call a path on the tree from a to b consists of vertices lie on the shortest simple path between a and b. And we say an edge is on the path if both its two endpoints is in the path, and an edge is adjacent to the path if exactly one endpoint of it is in the path.

Sometimes the devil will ask you to reverse every edge’s color on a path or adjacent to a path.

The king’s daughter, WJMZBMR, is also a cute loli, she is surprised by her father’s lolicon-like behavior. As she is concerned about the road-system’s status, sometimes she will ask you to tell there is how many black edge on a path.

Initially, every edges is white.

Input
The first line contains an integer T, denoting the number of the test cases.
For each test case, the first line contains an integer n, which is the size of the tree. The vertices be indexed from 1.
On the next n-1 lines, each line contains two integers a,b, denoting there is an edge between a and b.
The next line contains an integer Q, denoting the number of the operations.
On the next Q lines, each line contains three integers t,a,b. t=1 means we reverse every edge’s color on path a to b. t=2 means we reverse every edge’s color adjacent to path a to b. t=3 means we query about the number of black edge on path a to b.

T<=5.
n,Q<=10^5.

Output
For each t=3 operation, output the answer in one line.

Sample Input
1
10
2 1
3 1
4 1
5 1
6 5
7 4
8 3
9 5
10 6

10
2 1 6
1 3 8
3 8 10
2 3 4
2 10 8
2 4 10
1 7 6
2 7 3
2 1 4
2 10 10

Sample Output
3
```/*
hdu 4897 树链剖分(重轻链)

problem:

1.将u->v链上面的所有边的颜色翻转 (例：white -> black)
这个在线段树上很好处理,用个翻转标记,然后记录数量即可
2.将u->v链上面所有邻接的边翻转(边上只有一个点在链上面)
3.询问u->v上面有多少黑色的边

solve:

hhh-2016-08-18 21:18:55
*/
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#define lson  i<<1
#define rson  i<<1|1
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define key_val ch[ch[root][1]][0]
using namespace std;
const int maxn = 200100;
const int inf = 0x3f3f3f3f;
int top[maxn],fp[maxn],fa[maxn],dep[maxn],num[maxn],p[maxn];
int n;
struct Edge
{
int to,next;
} edge[maxn<<2];

void ini()
{
tot = 0,pos = 1;
}

{
}

void dfs1(int u,int pre,int d)
{
//    cout << u << " " <<pre <<" " <<d <<endl;
dep[u] = d;
fa[u] = pre,num[u] = 1;
for(int i = head[u]; ~i; i = edge[i].next)
{
int v = edge[i].to;
if(v != pre)
{
dfs1(v,u,d+1);
num[u] += num[v];
if(son[u] == -1 || num[v] > num[son[u]])
son[u] = v;
}
}
}

void getpos(int u,int sp)
{
top[u] = sp;
p[u] = pos++;
fp[p[u]] = u;
if(son[u] == -1)return ;
getpos(son[u],sp);
for(int i = head[u]; ~i ; i = edge[i].next)
{
int v = edge[i].to;
if(v != son[u] && v != fa[u])
getpos(v,v);
}
}

struct node
{
int l,r,mid;
int rev1,rev2;
int num;
} tree[maxn << 2];

void push_up(int i)
{
tree[i].num = tree[lson].num + tree[rson].num;
}

void build(int i,int l,int r)
{
tree[i].l = l,tree[i].r = r;
tree[i].mid=(l+r) >>1;
tree[i].rev1 = tree[i].rev2 = 0;
tree[i].num = 0;
if(l == r)
{
//        cout << fp[l] <<" " <<val[fp[l]]<<endl;
return;
}
build(lson,l,tree[i].mid);
build(rson,tree[i].mid+1,r);
}

void push_down(int i)
{
if(tree[i].rev1)
{
tree[i].rev1 = 0;
tree[lson].rev1 ^= 1,tree[lson].num = tree[lson].r-tree[lson].l+1-tree[lson].num;
tree[rson].rev1 ^= 1,tree[rson].num = tree[rson].r-tree[rson].l+1-tree[rson].num;;
}
if(tree[i].rev2)
{
tree[i].rev2 = 0;
tree[lson].rev2 ^= 1;
tree[rson].rev2 ^= 1;
}
}

void update_area(int i,int l,int r,int flag)
{
//    cout <<"l:"<< l <<" r:"<<r <<" min:"<< tree[i].Min<<endl;
if(l > r)
return ;
if(tree[i].l >= l && tree[i].r <= r)
{
if(flag == 1)
{
tree[i].num = tree[i].r-tree[i].l+1-tree[i].num;
tree[i].rev1 ^= 1;
}
else
tree[i].rev2 ^= 1;
return ;
}
push_down(i);
int mid = tree[i].mid;
if(r <= mid)
update_area(lson,l,r,flag);
else if(l > mid)
update_area(rson,l,r,flag);
else
{
update_area(lson,l,mid,flag);
update_area(rson,mid+1,r,flag);
}
push_up(i);
}

int query(int i,int l,int r,int flag)
{
if(l > r)
return 0;
if(tree[i].l >= l && tree[i].r <= r)
{
if(flag == 1)
return tree[i].num;
else
return tree[i].rev2;
}
push_down(i);
int mid = tree[i].mid;
if(r <= mid)
return query(lson,l,r,flag);
else if(l > mid)
return query(rson,l,r,flag);
else
return query(lson,l,mid,flag)+query(rson,mid+1,r,flag);
push_up(i);
}

void update_rev1(int u,int v)
{
int f1 = top[u],f2 = top[v];
while(f1 != f2)
{
if(dep[f1] < dep[f2])
{
swap(f1,f2),swap(u,v);
}
update_area(1,p[f1],p[u],1);
u = fa[f1],f1 = top[u];
}
if(dep[u] > dep[v]) swap(u,v);
update_area(1,p[son[u]],p[v],1);
}

void update_rev2(int u,int v)
{
int f1 = top[u],f2 = top[v];
//    cout << u << " " <<v<<endl;
while(f1 != f2)
{
if(dep[f1] < dep[f2])
{
swap(f1,f2),swap(u,v);
}
update_area(1,p[f1],p[u],2);
int par = fa[f1];
if(son[par] == f1) update_area(1,p[f1],p[f1],1);
if(son[u] != -1)  update_area(1,p[son[u]],p[son[u]],1);
u = fa[f1],f1 = top[u];
}
if(dep[u] > dep[v]) swap(u,v);

update_area(1,p[u],p[v],2);
int par = fa[u];
//    cout <<par <<" "<< son[v] <<endl;
if(son[par] == u && par > 0) update_area(1,p[u],p[u],1);
if(son[v] != -1)  update_area(1,p[son[v]],p[son[v]],1);
}

int Find(int u,int v)
{
//    cout <<"*********************************************************"<<endl;
int f1 = top[u],f2 = top[v];
int ans = 0;
//    cout << u << " " <<v<<endl;

while(f1 != f2)
{
if(dep[f1] < dep[f2])
{
swap(f1,f2),swap(u,v);
}
ans += query(1,p[f1]+1,p[u],1);
//        cout <<ans <<" " <<f1 <<" " <<u <<endl;
ans += query(1,p[fa[top[u]]],p[fa[top[u]]],2)^query(1,p[top[u]],p[top[u]],2)
^query(1,p[top[u]],p[top[u]],1);
//        cout <<ans<<" "<<fa[f1]<<" "<<u <<endl;
u = fa[f1],f1 = top[u];
}
if(dep[u] > dep[v]) swap(u,v);
//    cout << query(1,p[u]+1,p[v],1) <<endl;
return ans+query(1,p[u]+1,p[v],1);
}

int main()
{
//    freopen("in.txt","r",stdin);
int T,cas = 1,op;
int a,b;
int m,u,v;
scanf("%d",&T);
while(T--)
{
ini();
scanf("%d",&n);
for(int i =1; i <n; i++)
{
scanf("%d%d",&u,&v);
}
dfs1(1,0,0);
getpos(1,1);
build(1,1,pos-1);
scanf("%d",&m);
for(int i = 1; i <= m; i++)
{
scanf("%d%d%d",&op,&a,&b);
//            cout << op <<" " <<a <<" " <<b<<endl;
if(op == 1)
{
if(a == b)
continue;
update_rev1(a,b);
}
else if(op == 2)
{
update_rev2(a,b);
}
else if(op == 3)
{
printf("%d\n",Find(a,b));
}
}
}
return 0;
}
```

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