Codeforces 570D Tree Requests
D. Tree Requests
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 the n - 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 to 1.
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".
Codeforces Round #316 (Div. 2)
前天晚上做的题，昨天浪了一天，没来得及写
意思是有一棵树，每个节点都有一个字母
给出一组数(v,h)
询问 v的孩子中，到root 1的深度为h的所有节点的字母能不能排成一个对称的字符串 即出现奇数次的字母个数<=1个
当时不会做啊 怎么想都是m*n的方法 比如 用vecotr[u]保存所有u的孩子的节点编号 然后再做 但是保存孩子节点编号就是一个n^2的过程。。
今天看了题解 是用到了dfs序
这样 u的dfs序是(1,10)
那么dfs序在(1,10)之间的 都是u的孩子！
o(n)的时间可以预处理节点是否有关系
然后有两个方法
首先一个是可以用f[u][i]表示深度u,字符为i的dfs序
而且f[u][i]中的元素是递增的，可以二分
比如询问节点x 深度y
可以在f[u][0-25]中二分找到节点x的dfs序对应的下标
两个下标之差即为节点x 深度y 字符为i的节点个数 判断是否为奇数个 统计得出答案
交上去 1980ms 差点tle。。。
#include<bits/stdc++.h>
using namespace std;
vector<int>f[555555][27];
vector<int>g[555555];
int m,n;
char s[555555];
int tim,fst[555555],nxt[555555];
int x,y;
int d[555555];
void dfs(int u,int dis)
{
tim++;
fst[u]=tim;
d[u]=dis;
f[dis][s[u]-'a'].push_back(tim);
for(unsigned int i=0;i<g[u].size();i++)
dfs(g[u][i],dis+1);
nxt[u]=tim;
}
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 ti=1;ti<=n;ti++)
{
int has=0;
scanf("%d%d",&x,&y); //root x deepth y
int l=fst[x],r=nxt[x];
for(int i=0;i<26;i++)
{
int num = upper_bound(f[y][i].begin(),f[y][i].end(),r)-lower_bound(f[y][i].begin(),f[y][i].end(),l);
if(num%2==1)
has++;
}
if(has<2)
printf("Yes\n");
else
printf("No\n");
}
return 0;
}
还有一种方法
对于深度 直接保存所有的节点
利用xor判断节点出现次数
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
判断该值二进制每一位是否为1 为1表示该位对应的字母出现了奇数次 从而得出答案 只有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\n");
else
printf("No\n");
}
return 0;
}
