description
analysis
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用棧維護一下樹上路徑未匹配的左括號,然後在樹上找右括號匹配,設爲節點的貢獻,是答案
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爲左括號可以直接繼承父節點的信息,爲右括號且棧非空則可以匹配,貢獻值是棧頂左括號的父節點的貢獻
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這個其實就是當前子序列可以拼上左括號父親的序列,然後每一位的答案就是父節點的答案加上當前點的貢獻
code
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 500005
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)
#define rep(i,a) for (reg i=las[a];i;i=nex[i])
using namespace std;
ll las[MAXN],nex[MAXN],tov[MAXN];
ll f[MAXN],g[MAXN],fa[MAXN],stack[MAXN];
char s[MAXN];
ll n,tot,top,ans;
inline ll read()
{
ll x=0,f=1;char ch=getchar();
while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
return x*f;
}
inline void link(ll x,ll y){nex[++tot]=las[x],las[x]=tot,tov[tot]=y;}
inline void dfs(ll x)
{
ll tmp=0;
if (s[x]=='(')stack[++top]=x;
else if (top)tmp=stack[top],f[x]=f[fa[tmp]]+1,--top;
g[x]=g[fa[x]]+f[x],ans^=x*g[x];
rep(i,x)dfs(tov[i]);
if (tmp)stack[++top]=tmp;
else if (top)--top;
}
int main()
{
n=read(),scanf("%s",s+1);
fo(i,2,n)link(fa[i]=read(),i);
dfs(1),printf("%lld\n",ans);
return 0;
}