Codeforces Round #425 (Div. 2) Problem D Misha, Grisha and Underground (Codeforces 832D) - 树链剖分 - 树状数组
Misha and Grisha are funny boys, so they like to use new underground. The underground has n stations connected with n - 1 routes so that each route connects two stations, and it is possible to reach every station from any other.
The boys decided to have fun and came up with a plan. Namely, in some day in the morning Misha will ride the underground from station s to station f by the shortest path, and will draw with aerosol an ugly text "Misha was here" on every station he will pass through (including s and f). After that on the same day at evening Grisha will ride from station t to station f by the shortest path and will count stations with Misha's text. After that at night the underground workers will wash the texts out, because the underground should be clean.
The boys have already chosen three stations a, b and c for each of several following days, one of them should be station s on that day, another should be station f, and the remaining should be station t. They became interested how they should choose these stations s, f, t so that the number Grisha will count is as large as possible. They asked you for help.
The first line contains two integers n and q (2 ≤ n ≤ 105, 1 ≤ q ≤ 105) — the number of stations and the number of days.
The second line contains n - 1 integers p2, p3, ..., pn (1 ≤ pi ≤ n). The integer pi means that there is a route between stations pi and i. It is guaranteed that it's possible to reach every station from any other.
The next q lines contains three integers a, b and c each (1 ≤ a, b, c ≤ n) — the ids of stations chosen by boys for some day. Note that some of these ids could be same.
Print q lines. In the i-th of these lines print the maximum possible number Grisha can get counting when the stations s, t and f are chosen optimally from the three stations on the i-th day.
3 2
1 1
1 2 3
2 3 3
2 3
4 1
1 2 3
1 2 3
2
In the first example on the first day if s = 1, f = 2, t = 3, Misha would go on the route 1 
2, and Grisha would go on the route 3 1 2. He would see the text at the stations 1 and 2. On the second day, if s = 3, f = 2, t = 3, both boys would go on the route 3 1 2. Grisha would see the text at 3 stations.
In the second examle if s = 1, f = 3, t = 2, Misha would go on the route 1 2 3, and Grisha would go on the route 2 3 and would see the text at both stations.
题目大意 给定一个有n个点的树，有m个询问，每个询问给出3个点，你需要分别选出一个点作为s，f，t，如果将s到f的路径上的所有点权加1，计算t到f的路径上的点权和，问可能的点权和最大为多少。
显然树链剖分。
枚举6种可能，然后取个max。然后水完了。
然而比赛最后30分钟发现这特么是道水题，早知道该先写D题(内心崩溃...)。然后最后30分钟内20分钟写了出来，然后gdb挂了。。
然后在没gdb的情况下，今天上午把它改出来(不可思议)，然后发现，不会线段树区间lazy标记如何赋初值，树链剖分数组开始滥用。
另外线段树似乎容易被卡常，所以我就改用树状数组了（动态new点，能不TLE就怪了）。
（PS:其实这道题有一个很傻逼的$O(nlogn)$的做法，根本没必要树剖，直接求lca就行了，可惜当时没有想出来。。）
Code
/**
* Codeforces
* Problem#832D
* Accepted
* Time:1466ms
* Memory:14200k
*/
#include <iostream>
#include <cstdio>
#include <ctime>
#include <cmath>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <stack>
#ifndef WIN32
#define Auto "%lld"
#else
#define Auto "%I64d"
#endif
using namespace std;
typedef bool boolean;
const signed int inf = (signed)((1u << ) - );
const signed long long llf = (signed long long)((1ull << ) - );
const double eps = 1e-;
const int binary_limit = ;
#define smin(a, b) a = min(a, b)
#define smax(a, b) a = max(a, b)
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
template<typename T>
inline boolean readInteger(T& u){
char x;
int aFlag = ;
while(!isdigit((x = getchar())) && x != '-' && x != -);
if(x == -) {
ungetc(x, stdin);
return false;
}
if(x == '-'){
x = getchar();
aFlag = -;
}
for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
ungetc(x, stdin);
u *= aFlag;
return true;
} #define LL int
#define lowbit(x) (x & (-x)) typedef class IndexedTree {
public:
LL* a;
int s;
IndexedTree():a(NULL), s() { }
IndexedTree(int n):s(n) {
a = new LL[(n + )];
memset(a, , sizeof(LL) * (n + ));
} inline void add(int idx, LL val) {
for(; idx <= s; idx += lowbit(idx))
a[idx] += val;
} inline LL getSum(int idx) {
LL rt = ;
for(; idx; idx -= lowbit(idx))
rt += a[idx];
return rt;
}
}IndexedTree; typedef class SegmentableIndexedTree {
public:
IndexedTree s; // sum
IndexedTree us; // unique sum
stack< pair<int, int> > bs; SegmentableIndexedTree() { }
SegmentableIndexedTree(int n) {
s = IndexedTree(n + );
us = IndexedTree(n + );
} inline void add(int l, int r, LL val) {
s.add(l, val), us.add(l, l * val);
s.add(r + , -val), us.add(r + , -(r + ) * val);
if(val > )
bs.push(pair<int, int>(l, r));
} inline LL getSum(int idx) {
return (idx + ) * s.getSum(idx) - us.getSum(idx);
} inline LL getSum(int l, int r) {
return getSum(r) - getSum(l - );
} inline void reset() {
pair<int, int> p;
while(!bs.empty()) {
p = bs.top();
bs.pop();
add(p.first, p.second, -);
}
}
}SegmentableIndexedTree; const int maxn = 1e5 + ;
int n, m;
vector<int> g[maxn]; inline void init() {
readInteger(n);
readInteger(m);
for(int i = , x; i < n; i++) {
readInteger(x);
g[x].push_back(i + );
}
} int cnt = ;
int visitID[maxn], size[maxn], fa[maxn];
int zson[maxn], dep[maxn], top[maxn];
SegmentableIndexedTree st; void dfs1(int node, int last) {
dep[node] = dep[last] + ;
size[node] = ;
fa[node] = last;
int maxs = , maxid = ;
for(int i = ; i < (signed)g[node].size(); i++) {
int& e = g[node][i];
if(e == last) continue;
dfs1(e, node);
size[node] += size[e];
if(size[e] > maxs) maxs = size[e], maxid = e;//if(siz[son[u]] < siz[v]) son[u] = v;
}
zson[node] = maxid;
} void dfs2(int node, int last, boolean iszson) {
top[node] = (iszson) ? (top[last]) : (node);
visitID[node] = ++cnt;
if(zson[node] != ) dfs2(zson[node], node, true);
for(int i = ; i < (signed)g[node].size(); i++) {
int& e = g[node][i];
if(e == last || e == zson[node]) continue;
dfs2(e, node, false);
}
} void update(int a, int b){
// cout << st.query(st.root, 1, n, 1, n) << endl;
while(top[a] != top[b]){
int& d = (dep[top[a]] > dep[top[b]]) ? (a) : (b);
st.add(visitID[top[d]], visitID[d], );
d = fa[top[d]];
}
if(visitID[a] > visitID[b]) swap(a, b);
st.add(visitID[a], visitID[b], );
// cout << st.query(st.root, 1, n, 1, n) << endl;
} int query(int a, int b) {
int ret = ;
while(top[a] != top[b]){
int& d = (dep[top[a]] > dep[top[b]]) ? (a) : (b);
ret += st.getSum(visitID[top[d]], visitID[d]);
d = fa[top[d]];
}
if(visitID[a] > visitID[b]) swap(a, b);
ret += st.getSum(visitID[a], visitID[b]);
// cout << ret << endl;
// cout << "============================" << endl;
return ret;
} int query(int s, int t, int f) {
st.reset();
update(s, f);
return query(t, f);
} inline void solve() {
int a, b, c;
dfs1(, );
dfs2(, , false);
st = SegmentableIndexedTree(n);
// for(int i = 1; i <= n; i++)
// cout << zson[i] << " ";
while(m--) {
readInteger(a);
readInteger(b);
readInteger(c);
int res = query(a, b, c);
smax(res, query(b, a, c));
smax(res, query(b, c, a));
smax(res, query(b, a, c));
smax(res, query(c, a, b));
smax(res, query(c, b, a));
printf("%d\n", res);
}
} int main() {
init();
solve();
return ;
}
/**
* Codeforces
* Problem#832D
* Accepted
* Time:1466ms
* Memory:14200k
*/
#include <iostream>
#include <cstdio>
#include <ctime>
#include <cmath>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <stack>
#ifndef WIN32
#define Auto "%lld"
#else
#define Auto "%I64d"
#endif
using namespace std;
typedef bool boolean;
const signed int inf = (signed)((1u << ) - );
const signed long long llf = (signed long long)((1ull << ) - );
const double eps = 1e-;
const int binary_limit = ;
#define smin(a, b) a = min(a, b)
#define smax(a, b) a = max(a, b)
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
template<typename T>
inline boolean readInteger(T& u){
char x;
int aFlag = ;
while(!isdigit((x = getchar())) && x != '-' && x != -);
if(x == -) {
ungetc(x, stdin);
return false;
}
if(x == '-'){
x = getchar();
aFlag = -;
}
for(u = x - ''; isdigit((x = getchar())); u = (u << ) + (u << ) + x - '');
ungetc(x, stdin);
u *= aFlag;
return true;
} #define LL int
#define lowbit(x) (x & (-x)) typedef class IndexedTree {
public:
LL* a;
int s;
IndexedTree():a(NULL), s() { }
IndexedTree(int n):s(n) {
a = new LL[(n + )];
memset(a, , sizeof(LL) * (n + ));
} inline void add(int idx, LL val) {
for(; idx <= s; idx += lowbit(idx))
a[idx] += val;
} inline LL getSum(int idx) {
LL rt = ;
for(; idx; idx -= lowbit(idx))
rt += a[idx];
return rt;
}
}IndexedTree; typedef class SegmentableIndexedTree {
public:
IndexedTree s; // sum
IndexedTree us; // unique sum
stack< pair<int, int> > bs; SegmentableIndexedTree() { }
SegmentableIndexedTree(int n) {
s = IndexedTree(n + );
us = IndexedTree(n + );
} inline void add(int l, int r, LL val) {
s.add(l, val), us.add(l, l * val);
s.add(r + , -val), us.add(r + , -(r + ) * val);
if(val > )
bs.push(pair<int, int>(l, r));
} inline LL getSum(int idx) {
return (idx + ) * s.getSum(idx) - us.getSum(idx);
} inline LL getSum(int l, int r) {
return getSum(r) - getSum(l - );
} inline void reset() {
pair<int, int> p;
while(!bs.empty()) {
p = bs.top();
bs.pop();
add(p.first, p.second, -);
}
}
}SegmentableIndexedTree; const int maxn = 1e5 + ;
int n, m;
vector<int> g[maxn]; inline void init() {
readInteger(n);
readInteger(m);
for(int i = , x; i < n; i++) {
readInteger(x);
g[x].push_back(i + );
}
} int cnt = ;
int visitID[maxn], size[maxn], fa[maxn];
int zson[maxn], dep[maxn], top[maxn];
SegmentableIndexedTree st; void dfs1(int node, int last) {
dep[node] = dep[last] + ;
size[node] = ;
fa[node] = last;
int maxs = , maxid = ;
for(int i = ; i < (signed)g[node].size(); i++) {
int& e = g[node][i];
if(e == last) continue;
dfs1(e, node);
size[node] += size[e];
if(size[e] > maxs) maxs = size[e], maxid = e;//if(siz[son[u]] < siz[v]) son[u] = v;
}
zson[node] = maxid;
} void dfs2(int node, int last, boolean iszson) {
top[node] = (iszson) ? (top[last]) : (node);
visitID[node] = ++cnt;
if(zson[node] != ) dfs2(zson[node], node, true);
for(int i = ; i < (signed)g[node].size(); i++) {
int& e = g[node][i];
if(e == last || e == zson[node]) continue;
dfs2(e, node, false);
}
} void update(int a, int b){
// cout << st.query(st.root, 1, n, 1, n) << endl;
while(top[a] != top[b]){
int& d = (dep[top[a]] > dep[top[b]]) ? (a) : (b);
st.add(visitID[top[d]], visitID[d], );
d = fa[top[d]];
}
if(visitID[a] > visitID[b]) swap(a, b);
st.add(visitID[a], visitID[b], );
// cout << st.query(st.root, 1, n, 1, n) << endl;
} int query(int a, int b) {
int ret = ;
while(top[a] != top[b]){
int& d = (dep[top[a]] > dep[top[b]]) ? (a) : (b);
ret += st.getSum(visitID[top[d]], visitID[d]);
d = fa[top[d]];
}
if(visitID[a] > visitID[b]) swap(a, b);
ret += st.getSum(visitID[a], visitID[b]);
// cout << ret << endl;
// cout << "============================" << endl;
return ret;
} int query(int s, int t, int f) {
st.reset();
update(s, f);
return query(t, f);
} inline void solve() {
int a, b, c;
dfs1(, );
dfs2(, , false);
st = SegmentableIndexedTree(n);
// for(int i = 1; i <= n; i++)
// cout << zson[i] << " ";
while(m--) {
readInteger(a);
readInteger(b);
readInteger(c);
int res = query(a, b, c);
smax(res, query(b, a, c));
smax(res, query(b, c, a));
smax(res, query(b, a, c));
smax(res, query(c, a, b));
smax(res, query(c, b, a));
printf("%d\n", res);
}
} int main() {
init();
solve();
return ;
}
Codeforces Round #425 (Div. 2) Problem D Misha, Grisha and Underground (Codeforces 832D) - 树链剖分 - 树状数组的更多相关文章
- 【 Codeforces Round #425 (Div. 2) D】Misha, Grisha and Underground
[Link]:http://codeforces.com/contest/832/problem/D [Description] 给你一棵树; 然后给你3个点 让你把这3个点和点s,t,f对应; 然后 ...
- Codeforces Round #425 (Div. 2) Problem C Strange Radiation (Codeforces 832C) - 二分答案 - 数论
n people are standing on a coordinate axis in points with positive integer coordinates strictly less ...
- Codeforces Round #425 (Div. 2) Problem B Petya and Exam (Codeforces 832B) - 暴力
It's hard times now. Today Petya needs to score 100 points on Informatics exam. The tasks seem easy ...
- Codeforces Round #425 (Div. 2) Problem A Sasha and Sticks (Codeforces 832A)
It's one more school day now. Sasha doesn't like classes and is always bored at them. So, each day h ...
- Codeforces Round #425 (Div. 2) Misha, Grisha and Underground（LCA）
Misha, Grisha and Underground time limit per test 2 seconds memory limit per test 256 megabytes inpu ...
- Codeforces Round #716 (Div. 2), problem: (B) AND 0, Sum Big位运算思维
& -- 位运算之一,有0则0 原题链接 Problem - 1514B - Codeforces 题目 Example input 2 2 2 100000 20 output 4 2267 ...
- Codeforces Round #425 (Div. 2))——A题&&B题&&D题
A. Sasha and Sticks 题目链接:http://codeforces.com/contest/832/problem/A 题目意思:n个棍,双方每次取k个,取得多次数的人获胜,Sash ...
- Codeforces Round #425 (Div. 2) - D
题目链接:http://codeforces.com/contest/832/problem/D 题意:给定一棵n个点的树,然后给你q个询问,每个询问为三元组(a,b,c),问你从这三个点中选取一个作 ...
- Codeforces Round #425 (Div. 2) D.Misha, Grisha and Underground
我奇特的脑回路的做法就是 树链剖分 + 树状数组 树状数组是那种 区间修改,区间求和,还有回溯的 当我看到别人写的是lca,直接讨论时,感觉自己的智商收到了碾压... #include<cmat ...
随机推荐
- web.xml文件的简单说明
在javaEE提供的tutorial中的hello1中的web.xml文件写到: <?xml version="1.0" encoding="UTF-8" ...
- 海量交通大数据应用平台MTDAP_nchang的经验记录
WRONGTYPE Operation against a key holding the wrong kind of value 根本的就是redis同一个key的value值前后类型不一致,比如最 ...
- VScode编辑器个性化配置
一.设置方法 “文件” - > “首选项” -> "设置" 二.字体大小和缩进 "editor.tabSize": 2, "editor. ...
- Springboot整合Mybatis 之分页插件使用
1: 引入jar包 <!-- 引入MyBatis分页插件--> <dependency> <groupId>com.github.pagehelper</gr ...
- 三 js语句
/** * Created by Administrator on 2017/12/24. * 1.顺序语句 * 2.选择语句 if else swtich case * 3.循环语句 for whi ...
- uva 12222 Mountain Road
题意: 有一个单行道,两个方向都有车在等待.给出每个车的方向以及到达的时间以及走完这段路所需要的时间. 为了防止车祸,同向两车通过任一点的时间间隔不得小于10s. 求最后一辆车离开时刻的最小值. 思路 ...
- NSOperation、NSOperationQueue(II)
NSOperationQueue 控制串行执行.并发执行 NSOperationQueue 创建的自定义队列同时具有串行.并发功能 这里有个关键属性 maxConcurrentOperationCou ...
- Python OS模块常用功能 中文图文详解
一.Python OS模块介绍 OS模块简单的来说它是一个Python的系统编程的操作模块,可以处理文件和目录这些我们日常手动需要做的操作. 可以查看OS模块的帮助文档: >>> i ...
- eclipse xml 文件添加注解快捷键
eclipse xml 文件注解快捷键: <!-- --> Ctrl + shift + / 添加注解 Ctrl + shift + \ 取消注解
- django 常用方法总结 < 手写分页-上传头像-redis缓存，排行 ...>
1.不使用自带模块<Paginator>的手写分页功能views.pydef post_list(request): page = request.GET.get('page', 1) # ...