description

analysis

• 用栈维护一下树上路径未匹配的左括号，然后在树上找右括号匹配，设\(f[i]\)为\(i\)节点的贡献，\(g[i]\)是答案
• 为左括号可以直接继承父节点的信息，为右括号且栈非空则可以匹配，贡献值是栈顶左括号的父节点的贡献\(+1\)
• 这个其实就是当前子序列可以拼上左括号父亲的序列，然后每一位的答案就是父节点的答案加上当前点的贡献

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;
}

原文地址：https://www.cnblogs.com/horizonwd/p/12051481.html