【GDOI2017模拟11.2】相位幻击

Description

Input

Output

对于每个询问输出答案

Sample Input

3
1 2
1 3
1 2 3
2
C 1 1
Q 2 3

Sample Output

1

Solution

树链剖分裸题
你只需要知道一棵树的子树的dfs序是连续的就行了

Code

#include<cstdio>#include<algorithm>#include<cmath>#define fo(i,a,b) for(int i=a;i<=b;i++)#define N 1001000using namespace std;int last[N],next[N],to[N],tot=0,dfn[N],son[N],fa[N],dp[N],top[N],tal[N],a[N],dt[N],lz[N],size[N],an,n;void putin(int x,int y){    next[++tot]=last[x];last[x]=tot;to[tot]=y;}void dg1(int x){    size[x]=1;    for(int i=last[x];i;i=next[i])    if(dp[to[i]]==0) dp[to[i]]=dp[x]+1,fa[to[i]]=x,dg1(to[i]),size[x]+=size[to[i]],son[x]=size[to[i]]>size[son[x]]?to[i]:son[x];}void dg2(int x){    dfn[x]=++tot;if(son[x]) top[son[x]]=top[x],dg2(son[x]);    for(int i=last[x];i;i=next[i])    if(to[i]!=son[x]&&dp[to[i]]>dp[x]) top[to[i]]=to[i],dg2(to[i]);    tal[x]=tot;}void down(int v,int i,int j){    if(lz[v]==0) return;    int m=(i+j)/2;    if((m-i+1)%2==1)dt[v*2]^=lz[v];    if((j-m)%2==1)dt[v*2+1]^=lz[v];    lz[v*2]^=lz[v];lz[v*2+1]^=lz[v];lz[v]=0;}void change(int v,int i,int j,int x,int y,int z){    if(i==x&&j==y) {lz[v]^=z;if((j-i+1)%2==1)dt[v]^=z;return;}    int m=(i+j)/2;down(v,i,j);    if(y<=m) change(v*2,i,m,x,y,z);    else if(x>m) change(v*2+1,m+1,j,x,y,z);         else change(v*2,i,m,x,m,z),change(v*2+1,m+1,j,m+1,y,z);    dt[v]=dt[v*2]^dt[v*2+1];}void get(int v,int i,int j,int x,int y){    if(i==x&&j==y) {an^=dt[v];return;}    int m=(i+j)/2;down(v,i,j);    if(y<=m) get(v*2,i,m,x,y);    else if(x>m) get(v*2+1,m+1,j,x,y);         else get(v*2,i,m,x,m),get(v*2+1,m+1,j,m+1,y);    dt[v]=dt[v*2]^dt[v*2+1];}int lct(int x,int y){    int bz=1;an=0;    while(x!=y)    {        int f1=top[x],f2=top[y];        if(f1!=f2)        {            if(dp[f1]>=dp[f2]) get(1,1,n,dfn[f1],dfn[x]),x=fa[f1];            else get(1,1,n,dfn[f2],dfn[y]),y=fa[f2];        }        else        {            if(dp[x]<dp[y]) get(1,1,n,dfn[x],dfn[y]);else get(1,1,n,dfn[y],dfn[x]);            bz=0;break;        }    }    if(x!=0&&x==y&&bz) get(1,1,n,dfn[x],dfn[y]);    return an;}int main(){    freopen("phase.in","r",stdin);freopen("phase.out","w",stdout);    scanf("%d",&n);    fo(i,1,n-1)    {        int x,y;scanf("%d%d",&x,&y);        putin(x,y);putin(y,x);    }    dp[1]=top[1]=1;tot=0;dg1(1);dg2(1);    fo(i,1,n)     scanf("%d",&a[i]),change(1,1,n,dfn[i],dfn[i],a[i]);    int ac;    for(scanf("%d\n",&ac);ac;ac--)    {        int x,y;char ch;scanf("%c %d %d\n",&ch,&x,&y);        if(ch=='C') change(1,1,n,dfn[x],tal[x],y);        else printf("%d\n",lct(x,y));    }}