cf570d tree requests(dsu on tree)


dsu on tree————优雅的暴力

思路

首先重排能否构成回文串的条件显然是出现奇数次的字母只有1个或没有

所以联想到异或,最后数中为一的二进制位只有1个或没有即可

注意根节点深度为0

代码

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#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
int u[500100<<1],v[500100<<1],fir[500100],nxt[500100<<1],cnt,val[500100],w_p[500100],sz[500100],vis[500100],heason[500100],ans[500100],dep[500100],n,m;
struct {
int b,pid;
};
vector<ansNode> Vec[500100];
int calbit(int x){
int ans=0;
while(x){
ans++;
x&=(x-1);
}
return ans;
}
void change(int u,int f){
val[dep[u]]^=(1<<w_p[u]);
for(int i=fir[u];i;i=nxt[i]){
if(vis[v[i]]||v[i]==f)
continue;
change(v[i],u);
}
}
void addedge(int ui,int vi){
++cnt;
u[cnt]=ui;
v[cnt]=vi;
nxt[cnt]=fir[ui];
fir[ui]=cnt;
}
void dfs1(int u,int f){
sz[u]=1;
dep[u]=dep[f]+1;
for(int i=fir[u];i;i=nxt[i]){
if(v[i]==f)
continue;
dfs1(v[i],u);
sz[u]+=sz[v[i]];
if(heason[u]==0||sz[heason[u]]<sz[v[i]])
heason[u]=v[i];
}
}
void dfs2(int u,int f,int islight){
for(int i=fir[u];i;i=nxt[i]){
if(v[i]==f||v[i]==heason[u])
continue;
dfs2(v[i],u,1);
}
if(heason[u])
dfs2(heason[u],u,0),vis[heason[u]]=true;
change(u,f);
for(int i=0;i<Vec[u].size();i++)
ans[Vec[u][i].pid]=(calbit(val[Vec[u][i].b])==1||calbit(val[Vec[u][i].b])==0);
if(heason[u])
vis[heason[u]]=false;
if(islight)
change(u,f);
}
int main(){

// int x;
// scanf("%d",&x);
// printf("%dn",calbit(x));
// }
dep[0]=0;
scanf("%d %d",&n,&m);
for(int i=2;i<=n;i++){
int x;
scanf("%d",&x);
addedge(x,i);
addedge(i,x);
}
for(int i=1;i<=n;i++){
char c=getchar();
while(c<'a'||c>'z')
c=getchar();
w_p[i]=c-'a';
}
for(int i=1;i<=m;i++){
int x,y;
scanf("%d %d",&x,&y);
Vec[x].push_back((ansNode){y,i});
}
dfs1(1,0);
dfs2(1,0,0);
for(int i=1;i<=m;i++)
printf("%sn",(ans[i])?"Yes":"No");
return 0;
}