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Codeforces Round #316 (Div. 2) D. Tree Requests 樹離線在線算法

編輯：關於C++

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

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.

題意，給出一個樹，要求一些詢問，某個結點的子樹中的第q層能否形成回文串。

1.狀態壓縮，由於，只有26個字符，所以用一個int存每個字符的奇偶數（只需要知道奇偶性，下面也只需要用奇偶性），壓縮內存。

2.如何判定回文，只要a-z的字符最多有一個字符出現奇數次，就可以形成回文串。也就是（x&(-x)),去掉最後的一個1後，還是非0說明存在2個1以上，也就不能構成回文

方法一:先來說下離線的做法，先把所有的詢問讀入，按dfs搜索的順序，每個詢問在進入前先異或上這層的值，出這個結點時，再異或上這層的值，先前該層的值異或兩次抵消了，所以就是這個結點下的子樹的該層的值了。復雜度為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 p[N];
vector 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
);
            else
                printf(Yes
);
        }
    }
    return 0;
}

方法二：在線做法。先把樹形結構轉化成線形結構，也就是進入的時候標記一下，出來的時候標記一下，就可以轉化成為一個線形增大的值。每個結點，只會影響到它所在層的值，所以，先預處理出每個結點在它的層中形成的狀態存起來。然後，由於是要求結點v的子樹，所以只要求出在s[v] e[v]之間的值，用二分的方法，找出起始結點的位置，和離開結點的位置用結束時的狀態異或開始的狀態，就可以求出來這個結點子樹的狀態了。復雜度由於要二分查找，為o(n * log(n);但離線的做法，更加有意義。

#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 p[N];
vector 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
);
            else
                printf(Yes
);
        }
    }
    return 0;
}