Roman planted a tree consisting of n vertices. Each vertex contains a lowercase English letter. Vertex 1 is the root of the tree, each of then - 1 remaining vertices has a parent in the tree. Vertex is connected with its parent by an edge. The parent of vertex i is vertex pi, the parent index is always less than the index of the vertex (i.e., pi < i).
The depth of the vertex is the number of nodes on the path from the root to v along the edges. In particular, the depth of the root is equal to1.
We say that vertex u is in the subtree of vertex v, if we can get from u to v, moving from the vertex to the parent. In particular, vertex v is in its subtree.
Roma gives you m queries, the i-th of which consists of two numbers vi, hi. Let's consider the vertices in the subtree vi located at depth hi. Determine whether you can use the letters written at these vertices to make a string that is a palindrome. The letters that are written in the vertexes, can be rearranged in any order to make a palindrome, but all letters should be used.
The first line contains two integers n, m (1 ≤ n, m ≤ 500 000) — the number of nodes in the tree and queries, respectively.
The following line contains n - 1 integers p2, p3, ..., pn — the parents of vertices from the second to the n-th (1 ≤ pi < i).
The next line contains n lowercase English letters, the i-th of these letters is written on vertex i.
Next m lines describe the queries, the i-th line contains two numbers vi, hi (1 ≤ vi, hi ≤ n) — the vertex and the depth that appear in the i-th query.
Print m lines. In the i-th line print "Yes" (without the quotes), if in the i-th query you can make a palindrome from the letters written on the vertices, otherwise print "No" (without the quotes).
6 5 1 1 1 3 3 zacccd 1 1 3 3 4 1 6 1 1 2
Yes No Yes Yes Yes
String s is a palindrome if reads the same from left to right and from right to left. In particular, an empty string is a palindrome.
Clarification for the sample test.
In the first query there exists only a vertex 1 satisfying all the conditions, we can form a palindrome "z".
In the second query vertices 5 and 6 satisfy condititions, they contain letters "с" and "d" respectively. It is impossible to form a palindrome of them.
In the third query there exist no vertices at depth 1 and in subtree of 4. We may form an empty palindrome.
In the fourth query there exist no vertices in subtree of 6 at depth 1. We may form an empty palindrome.
In the fifth query there vertices 2, 3 and 4 satisfying all conditions above, they contain letters "a", "c" and "c". We may form a palindrome "cac".
题意：
给定n个点的树。m个询问
以下n-1个数给出每一个点的父节点。1是root
每一个点有一个字母
以下n个小写字母给出每一个点的字母。
以下m行给出询问：
询问形如 (u, deep) 问u点的子树中，距离根的深度为deep的全部点的字母是否能在随意排列后组成回文串，能输出Yes.
思路：dfs序。给点又一次标号，dfs进入u点的时间戳记为l[u], 离开的时间戳记为r[u], 这样对于某个点u，他的子树节点相应区间都在区间 [l[u], r[u]]内。
把距离根深度同样的点都存到vector里 D[i] 表示深度为i的全部点，在dfs时能够顺便求出。
把询问按深度排序,query[i]表示全部深度为i的询问。
接下来依照深度一层层处理。
对于第i层，把全部处于第i层的节点都更新到26个树状数组上。
然后处理询问，直接查询树状数组上有多少种字母是奇数个的，显然奇数个字母的种数要<=1
处理完第i层。就把树状数组逆向操作，相当于清空树状数组
注意的一个地方就是 询问的深度是随意的，也就是说可能超过实际树的深度。也可能比当前点的深度小。。所以须要初始化一下答案。
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <vector>
#include <string>
#include <time.h>
#include <math.h>
#include <iomanip>
#include <queue>
#include <stack>
#include <set>
#include <map>
const int inf = 1e9;
const double eps = 1e-8;
const double pi = acos(-1.0);
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if (c = getchar(), c == EOF) return 0;
while (c != '-' && (c<'0' || c>'9')) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x < 0) { putchar('-'); x = -x; }
if (x > 9) pt(x / 10);
putchar(x % 10 + '0');
}
using namespace std;
const int N = 5e5 + 100;
typedef long long ll;
typedef pair<int, int> pii;
struct BIT {
int c[N], maxn;
void init(int n) { maxn = n; memset(c, 0, sizeof c); }
inline int Lowbit(int x) { return x&(-x); }
void change(int i, int x)//i点增量为x
{
while (i <= maxn)
{
c[i] += x;
i += Lowbit(i);
}
}
int sum(int x) {//区间求和 [1,x]
int ans = 0;
for (int i = x; i >= 1; i -= Lowbit(i))
ans += c[i];
return ans;
}
int query(int l, int r) {
return sum(r) - sum(l - 1);
}
}t[26];
int n, m;
char s[N];
vector<int>G[N], D[N];
int l[N], r[N], top;
vector<pii>query[N];
bool ans[N];
void dfs(int u, int fa, int dep) {
D[dep].push_back(u);
l[u] = ++top;
for (auto v : G[u])
if (v != fa)dfs(v, u, dep + 1);
r[u] = top;
}
int main() {
rd(n); rd(m);
fill(ans, ans + m + 10, 1);
for (int i = 0; i < 26; i++) t[i].init(n);
for (int i = 2, u; i <= n; i++)rd(u), G[u].push_back(i);
top = 0;
dfs(1, 1, 1);
scanf("%s", s + 1);
for (int i = 1, u, v; i <= m; i++) {
rd(u); rd(v); query[v].push_back(pii(u, i));
}
for (int i = 1; i <= n; i++)
{
if (D[i].size() == 0)break;
for (auto v : D[i]) t[s[v] - 'a'].change(l[v], 1);
for (pii Q : query[i])
{
int ou = 0;
for (int j = 0; j < 26; j++)
{
if (t[j].query(l[Q.first], r[Q.first]))
ou += t[j].query(l[Q.first], r[Q.first]) & 1;
}
ans[Q.second] = ou <= 1;
}
for (auto v : D[i]) t[s[v] - 'a'].change(l[v], -1);
}
for (int i = 1; i <= m; i++)ans[i] ?
puts("Yes") : puts("No");
return 0;
}
另一种方法
对于深度 直接保存全部的节点
利用xor推断节点出现次数
这样二分到两个下标l,r时
myxor[r]^myxor[l-1]即为l+1到r的xor值
推断该值二进制每一位是否为1 为1表示该位相应的字母出现了奇数次 从而得出答案 仅仅有700ms左右
#include<bits/stdc++.h>
using namespace std;
vector<int>f[555555];
vector<int>g[555555];
vector<int> myxor[555555];
int m,n;
char s[555555];
int tim,fst[555555],nxt[555555];
int x,y;
int maxdeep=0;
int hsh[555555];
void dfs(int u,int dis)
{
tim++;
fst[u]=tim;
hsh[tim]=u;
f[dis].push_back(tim);
for(unsigned int i=0;i<g[u].size();i++)
dfs(g[u][i],dis+1);
nxt[u]=tim;
maxdeep=max(maxdeep,dis);
}
void work(int l,int r,int d,int &ans)
{
r--;
l--;
int ret;
if(r<0)
return;
ret=myxor[d][r];
if(l>=0)
ret^=myxor[d][l];
while(ret)
{
ans=ans+(ret&1);
ret>>=1;
}
}
int main()
{
scanf("%d%d",&m,&n);
for(int i=2;i<=m;i++)
{
scanf("%d",&x);
g[x].push_back(i);
}
for(int i=1;i<=m;i++)
{
scanf(" %c",&s[i]);
}
dfs(1,1);
for(int i=1;i<=maxdeep;i++)
{
myxor[i].push_back(1<<(s[hsh[f[i][0]]]-'a'));
for(int j=1;j<f[i].size();j++)
{
myxor[i].push_back((1<<(s[hsh[f[i][j]]]-'a'))^myxor[i][j-1]);
}
}
int has;
for(int ti=1;ti<=n;ti++)
{
has=0;
scanf("%d%d",&x,&y); //root x deepth y
int l =lower_bound(f[y].begin(),f[y].end(),fst[x])-f[y].begin();
int r =upper_bound(f[y].begin(),f[y].end(),nxt[x])-f[y].begin();
work(l,r,y,has);
if(has<2)
printf("Yes
");
else
printf("No
");
}
return 0;
}#include<bits/stdc++.h>
using namespace std;
vector<int>G[500009];
vector<int>Query[500009];
int dep[500009];
char str[500009];
int ans[500009];
int height[500009];
void dfs(int u,int d)
{
for(int v:Query[u]) ans[v]^=dep[height[v]];
for(int v:G[u]) dfs(v,d+1);
dep[d]^=(1<<(int)(str[u]-'a'));
for(int v:Query[u]) ans[v]^=dep[height[v]];
}
int main()
{
int n,m,x,y;
scanf("%d%d",&n,&m);
for(int i=2;i<=n;i++)
{
scanf("%d",&x);
G[x].push_back(i);
}
scanf("%s",(str+1));
for(int i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
Query[x].push_back(i);
height[i]=y;
}
dfs(1,1);
for(int i=1;i<=m;i++)
{
if(ans[i]&(ans[i]-1)) printf("No
");
else printf("Yes
");
}
return 0;
}