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  • codeforces 570 D. Tree Requests 树状数组+dfs搜索序

    链接:http://codeforces.com/problemset/problem/570/D


    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 vihi. 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 nm (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 vihi (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:

    告诉你一颗树的父子关系,1节点为根。再告诉你每一个点上的字母。

    问 v节点 子树(包含v节点)在第h行的全部节点的字母是否能组成回文串。


    做法:

    先用dfs 搜索 把全部节点标个左标号和右标号。 这样标号以后。每一个节点 用左标号 当自己 新的标号。 然后  子树全部节点 的新标号 肯定在 子树根节点的 左右标号之间。

    标号之后分层来做。

    每层  对每一个字母分别做统计。

    把该层全部节点 的 左标号 在树状数组中+1. 然后对于该层的全部询问 做 树状数组统计。(sum(rit[v])-sum(lft[v]-1))。   

    假设是奇数 说明这个 字母在查询的区间内 有奇数个。

    每一个查询  最多有一个奇数个的字母。否则不能构成回文串


    #include <iostream>
    #include<cstdio>
    #include<cstdlib>
    #include<cstring>
    #include<vector>
    #include<math.h>
    
    using namespace std;
    const int N = 500100;
    
    int f[N];
    vector<int> son[N];
    int id;
    int lft[N],rit[N];
    int deps[N]; 
    int ans[N];
    char str[N];
    vector<int> G[N];//深度
    vector<pair<int,int> > Q[N];
    void dfs(int nw,int dep)
    {
    	lft[nw]=id++;
    	deps[nw]=dep;
    	G[dep].push_back(nw);
    	for(int i=0;i<son[nw].size();i++)
    	{
    		int to=son[nw][i];
    		dfs(to,dep+1);
    	}
    	rit[nw]=id++;
    }
    
    int bit[2*N];
    
    int lowbit(int x)
    {
    	return x&(-x);
    }
    
    void add(int wei,int x)
    {
    
    	while(wei<=id)
    	{
    		bit[wei]+=x;
    		wei+=lowbit(wei);
    	}
    }
    
    int sum(int wei)
    {
    	if(wei==0)
    		return 0;
    	int sum=0;
    	while(wei>0)
    	{
    		sum+=bit[wei];
    		wei-=lowbit(wei);
    	}
    	return sum;
    }
    
    int main()
    {
    	int n,m;
    	while(scanf("%d%d",&n,&m)!=EOF)
    	{
    		for(int i=2;i<=n;i++)
    		{
    			scanf("%d",f+i);
    			son[f[i]].push_back(i);
    		}
    		scanf("%s",str+1);
    		id=1;
    		dfs(1,1);
    		int dep=1;
    		for(int i=1;i<=m;i++)
    		{
    			int vv,hh;
    			scanf("%d%d",&vv,&hh);
    			dep=max(hh,dep);
    			Q[hh].push_back(make_pair<int,int>(vv,i));
    		}
    
    		for(int i=1;i<=dep;i++)
    		{
    			for(int j=0;j<26;j++)
    			{
    				if(j==25)
    					int kkk=1;
    				for(int k=0;k<G[i].size();k++)//每一个节点
    				{
    					if(str[G[i][k]]-'a'==j)
    						add(lft[G[i][k]],1);
    				}
    				
    				for(int k=0;k<Q[i].size();k++)
    				{
    					int v=Q[i][k].first;
    					int ii=Q[i][k].second;
    					if((sum(rit[v])-sum(lft[v]-1))&1) ans[ii]++;
    				}
    
    				 
    				for(int k=0;k<G[i].size();k++)//每一个节点
    				{
    					if(str[G[i][k]]-'a'==j)
    						add(lft[G[i][k]],-1);
    				}
    				// printf("jj %d 
    ",j);
    			}
    
    			//printf("ii %d 
    ",i);
    		}
    
    		for(int i=1;i<=m;i++)
    		{
    			if(ans[i]<=1)
    				printf("Yes
    ");
    			else
    				printf("No
    ");
    		}
    
    	}
    	return 0;
    }




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  • 原文地址:https://www.cnblogs.com/claireyuancy/p/7199774.html
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