D. Tree Requests
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 depthhi.
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.
Input
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 thei-th query.
Output
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).
Sample test(s)
input
6 5 1 1 1 3 3 zacccd 1 1 3 3 4 1 6 1 1 2
output
Yes No Yes Yes Yes
Note
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，给出一个树，要求一些询问，某个结点的子树中的第q层能否形成回文串。
1.状态压缩，由于，只有26个字符，所以用一个int存每个字符的奇偶数（只需要知道奇偶性，下面也只需要用奇偶性），压缩内存。
2.如何判定回文，只要a-z的字符最多有一个字符出现奇数次，就可以形成回文串。也就是（x&(-x)),去掉最后的一个1后，还是非0说明存在2个1以上，也就不能构成回文
方法一:先来说下离线的做法，先把所有的询问读入，按dfs搜索的顺序，每个询问在进入前先异或上这层的值，出这个结点时，再异或上这层的值，先前该层的值异或两次抵消了，所以就是这个结点下的子树的该层的值了。复杂度为o(n);
#define N 500005
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,m,tree[N],t,v,h,ans[N],pre[N];
char str[N];
bool vis[N];
vector<int> p[N];
vector<pii> q[N];
int DFS(int f,int depth){
FI(q[f].size()){
ans[q[f][i].first] ^= pre[q[f][i].second];
}
FI(p[f].size()){
DFS(p[f][i],depth + 1);
}
pre[depth] ^= (1<<(str[f - 1] - 'a'));
FI(q[f].size()){
ans[q[f][i].first] ^= pre[q[f][i].second];
}
return 0;
}
int main()
{
while(S2(n,m)!=EOF)
{
FI(n + 1){
p[i].clear();
q[i].clear();
pre[i] = 0;
ans[i] = 0;
}
FI(n-1){
S(t);
p[t].push_back(i+2);
}
SS(str);
FI(m){
S2(v,h);
q[v].push_back(mp(i,h));
}
DFS(1,1);
FI(m){
if(ans[i] & (ans[i] - 1))
printf("No\n");
else
printf("Yes\n");
}
}
return 0;
}
方法二：在线做法。先把树形结构转化成线形结构，也就是进入的时候标记一下，出来的时候标记一下，就可以转化成为一个线形增大的值。每个结点，只会影响到它所在层的值，所以，先预处理出每个结点在它的层中形成的状态存起来。然后，由于是要求结点v的子树，所以只要求出在s[v] e[v]之间的值，用二分的方法，找出起始结点的位置，和离开结点的位置用结束时的状态异或开始的状态，就可以求出来这个结点子树的状态了。复杂度由于要二分查找，为o(n * log(n);但离线的做法，更加有意义。
#define N 500005
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,m,tree[N],t,v,h,ans,s[N],e[N],index;
char str[N];
bool vis[N];
vector<int> p[N];
vector<pii> dep[N];
int DFS(int f,int depth){
s[f] = index++;
FI(p[f].size()){
DFS(p[f][i],depth + 1);
}
dep[depth].push_back(mp(index,dep[depth].back().second^(1<<(str[f - 1] - 'a'))));
e[f] = index++;
return 0;
}
int main()
{
while(S2(n,m)!=EOF)
{
FI(n + 1){
p[i].clear();
dep[i].clear();
dep[i].push_back(mp(0,0));
}
FI(n-1){
S(t);
p[t].push_back(i+2);
}
SS(str);
index = 1;
DFS(1,1);
FI(m){
S2(v,h);
int l = lower_bound(dep[h].begin(),dep[h].end(),mp(s[v],-1)) - dep[h].begin() - 1;
int r = lower_bound(dep[h].begin(),dep[h].end(),mp(e[v],-1)) - dep[h].begin() - 1;
ans = dep[h][r].second ^ dep[h][l].second;
if(ans & (ans - 1))
printf("No\n");
else
printf("Yes\n");
}
}
return 0;
}
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