Codeforces 570D TREE REQUESTS dfs序+树状数组
题解链接:点击打开链接
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;
}
2 seconds
256 megabytes
standard input
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.
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.
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 570D TREE REQUESTS dfs序+树状数组的更多相关文章
- Codeforces 570D TREE REQUESTS dfs序+树状数组 异或
http://codeforces.com/problemset/problem/570/D Tree Requests time limit per test 2 seconds memory li ...
- poj3321-Apple Tree(DFS序+树状数组)
Apple Tree Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 36442 Accepted: 10894 Desc ...
- pku-3321 Apple Tree(dfs序+树状数组)
Description There is an apple tree outside of kaka's house. Every autumn, a lot of apples will grow ...
- POJ 3321 Apple Tree(dfs序树状数组)
http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=10486 题意:一颗有n个分支的苹果树,根为1,每个分支只有一个苹果,给出n- ...
- POJ 3321:Apple Tree(dfs序+树状数组)
题目大意:对树进行m次操作,有两类操作,一种是改变一个点的权值(将0变为1,1变为0),另一种为查询以x为根节点的子树点权值之和,开始时所有点权值为1. 分析: 对树进行dfs,将树变为序列,记录每个 ...
- CodeForces 570D - Tree Requests - [DFS序+二分]
题目链接:https://codeforces.com/problemset/problem/570/D 题解: 这种题,基本上容易想到DFS序. 然后,我们如果再把所有节点分层存下来,那么显然可以根 ...
- Codeforces Round #225 (Div. 1) C. Propagating tree dfs序+树状数组
C. Propagating tree Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/383/p ...
- Codeforces Round #225 (Div. 1) C. Propagating tree dfs序+ 树状数组或线段树
C. Propagating tree Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/383/p ...
- HDU 5293 Tree chain problem 树形dp+dfs序+树状数组+LCA
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5293 题意: 给你一些链,每条链都有自己的价值,求不相交不重合的链能够组成的最大价值. 题解: 树形 ...
随机推荐
- usb键鼠驱动分析【钻】
本文转载自:http://blog.csdn.net/orz415678659/article/details/9197859 一.鼠标 Linux下的usb鼠标驱动在/drivers/hid/usb ...
- Android.mk中添加宏定义【转】
本文转载自:http://blog.csdn.net/huangyabin001/article/details/38302021 在Boardconfig.mk 中添加一个 IS_FLAG := t ...
- linux udev 机制【转】
本文转载自:http://blog.csdn.net/yyt8yyt8/article/details/8020154 1. Linux的热插拔事件由kernel通过中断发现(比如,USB设备插入系统 ...
- DNS 隐蔽通道工具资料汇总
http://www.cnblogs.com/bonelee/p/7651746.html DNS隧道和工具 内含dns2tcp.iodine.dnscat2工具的简单使用说明 iodine工具的使用 ...
- Bone Collector(hdoj--2602--01背包)
Bone Collector Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) T ...
- JXOI2019 退役记
day0 考前一天在机房和RyeCatcher,还有高一数竞大佬wyt一起颓三国杀,被深深吸引无法自拔,所谓大考大浪,也算是缓解缓解压力 刷刷空间发现好多外地OIer都赶到江科了,萌生出去见一见我江西 ...
- C Linux read write function extension
前言 - 赠送 readn / writen Linux 上默认的 read 和 write 函数会被信号软中断. 且 read 和 write 函数中第三个参数 count #include < ...
- MyBatis输出执行的SQL到控制台
src\main\resources\application.properties 或者src\main\resources\application.yml 在你的application.proper ...
- 基于scrapy-redis组件的分布式爬虫
scrapy-redis组件安装 分布式实现流程 scrapy-redis组件安装 - 下载scrapy-redis组件:pip install scrapy-redis - 更改redis配置文件: ...
- 深入了解Token认证的来龙去脉
Token 是在服务端产生的,如果前端使用用户名/密码向服务端请求认证,服务端认证成功,那么在服务端会返回 Token 给前端.前端可以在每次请求的时候带上 Token 证明自己的合法地位. 不久 ...