HDU - 3966-Aragorn' Story(树链剖分+线段树)
链接:https://vjudge.net/problem/HDU-3966
题意:
Our protagonist is the handsome human prince Aragorn comes from The Lord of the Rings. One day Aragorn finds a lot of enemies who want to invade his kingdom. As Aragorn knows, the enemy has N camps out of his kingdom and M edges connect them. It is guaranteed that for any two camps, there is one and only one path connect them. At first Aragorn know the number of enemies in every camp. But the enemy is cunning , they will increase or decrease the number of soldiers in camps. Every time the enemy change the number of soldiers, they will set two camps C1 and C2. Then, for C1, C2 and all camps on the path from C1 to C2, they will increase or decrease K soldiers to these camps. Now Aragorn wants to know the number of soldiers in some particular camps real-time.
思路:
第一次写树链剖分, 直接上模板。
代码:
#include <iostream>
#include <memory.h>
#include <string>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <algorithm>
#include <map>
#include <queue>
#include <math.h>
#include <cstdio>
#include <set>
#include <iterator>
#include <cstring>
#include <assert.h>
using namespace std; typedef long long LL;
const int MAXN = 5e4+10;
vector<int> G[MAXN];
int Dis[MAXN], Fa[MAXN], Top[MAXN], Size[MAXN];
int Son[MAXN], Id[MAXN], Rk[MAXN];
int Seg[MAXN*4], A[MAXN], Add[MAXN*4];
int n, m, p;
int x, y;
int cnt = 0; void Init()
{
for (int i = 1;i <= n;i++)
G[i].clear();
memset(Seg, 0, sizeof(Seg));
memset(Son, 0, sizeof(Son));
memset(Add, 0, sizeof(Add));
cnt = 0;
} void Dfs1(int x, int u, int dep)
{
Dis[u] = dep;
Fa[u] = x;
Size[u] = 1;
for (int i = 0;i < G[u].size();i++)
{
int v = G[u][i];
if (v == x)
continue;
Dfs1(u, v, dep+1);
Size[u] += Size[v];
if (Size[v] > Size[Son[u]])
Son[u] = v;
}
} void Dfs2(int u, int top)
{
Top[u] = top;
Id[u] = ++cnt;
Rk[cnt] = u;
if (!Son[u])
return;
Dfs2(Son[u], top);
for (int i = 0;i < G[u].size();i++)
{
int v = G[u][i];
if (v != Son[u] && v != Fa[u])
Dfs2(v, v);
}
} void PushUp(int root)
{
Seg[root] = Seg[root<<1]+Seg[root<<1|1];
} void PushDown(int root, int l, int r)
{
if (Add[root])
{
int mid = (l+r)/2;
Add[root<<1] += Add[root];
Add[root<<1|1] += Add[root];
Seg[root<<1] += (mid-l+1)*Add[root];
Seg[root<<1|1] += (r-mid)*Add[root];
Add[root] = 0;
}
} void Build(int root, int l, int r)
{
if (l == r)
{
Seg[root] = A[Rk[l]];
return;
}
int mid = (l+r)/2;
Build(root<<1, l, mid);
Build(root<<1|1, mid+1, r);
PushUp(root);
} int Query(int root, int l, int r, int ql, int qr)
{
if (r < ql || qr < l)
return 0;
if (ql <= l && r <= qr)
return Seg[root];
PushDown(root, l, r);
int mid = (l+r)/2;
int res = 0;
res += Query(root<<1, l, mid, ql, qr);
res += Query(root<<1|1, mid+1, r, ql, qr);
PushUp(root);
return res;
} void Update(int root, int l, int r, int ql, int qr, int c)
{
if (r < ql || qr < l)
return;
if (ql <= l && r <= qr)
{
Seg[root] += (r-l+1)*c;
Add[root] += c;
return;
}
PushDown(root, l, r);
int mid = (l+r)/2;
Update(root<<1, l, mid, ql, qr, c);
Update(root<<1|1, mid+1, r, ql, qr, c);
PushUp(root);
} void UpdateLine(int l, int r, int c)
{
while (Top[l] != Top[r])
{
if (Dis[Top[l]] < Dis[Top[r]])
swap(l, r);
Update(1, 1, n, Id[Top[l]], Id[l], c);
l = Fa[Top[l]];
}
if (Id[l] < Id[r])
Update(1, 1, n, Id[l], Id[r], c);
else
Update(1, 1, n, Id[r], Id[l], c);
} int main()
{
// freopen("test.in", "r", stdin);
while (~scanf("%d%d%d", &n, &m, &p))
{
Init();
for (int i = 1;i <= n;i++)
scanf("%d", &A[i]);
for (int i = 1;i < n;i++)
{
scanf("%d%d", &x, &y);
G[x].push_back(y);
G[y].push_back(x);
}
Dfs1(0, 1, 1);
Dfs2(1, 1);
Build(1, 1, n);
// for (int i = 1;i <= n;i++)
// cout << Dis[i] << ' ';
// cout << endl;
// for (int i = 1;i <= n;i++)
// cout << Id[i] << ' ';
// cout << endl;
// for (int i = 1;i <= n;i++)
// cout << Top[i] << ' ' ;
// cout << endl;
char op[10];
int l, r, v, w;
while (p--)
{
scanf("%s", op);
if (op[0] == 'I')
{
scanf("%d%d%d", &l, &r, &v);
// cout << Top[l] << ' ' << Top[r] << endl;
UpdateLine(l, r, v);
}
else if (op[0] == 'D')
{
scanf("%d%d%d", &l, &r, &v);
v = -v;
UpdateLine(l, r, v);
}
else
{
scanf("%d", &w);
printf("%d\n", Query(1, 1, n, Id[w], Id[w]));
}
}
} return 0;
}
HDU - 3966-Aragorn' Story(树链剖分+线段树)的更多相关文章
- Aragorn's Story 树链剖分+线段树 && 树链剖分+树状数组
Aragorn's Story 来源:http://www.fjutacm.com/Problem.jsp?pid=2710来源:http://acm.hdu.edu.cn/showproblem.p ...
- HDU 2460 Network(双连通+树链剖分+线段树)
HDU 2460 Network 题目链接 题意:给定一个无向图,问每次增加一条边,问个图中还剩多少桥 思路:先双连通缩点,然后形成一棵树,每次增加一条边,相当于询问这两点路径上有多少条边,这个用树链 ...
- 【BZOJ-2325】道馆之战 树链剖分 + 线段树
2325: [ZJOI2011]道馆之战 Time Limit: 40 Sec Memory Limit: 256 MBSubmit: 1153 Solved: 421[Submit][Statu ...
- 【BZOJ2243】[SDOI2011]染色 树链剖分+线段树
[BZOJ2243][SDOI2011]染色 Description 给定一棵有n个节点的无根树和m个操作,操作有2类: 1.将节点a到节点b路径上所有点都染成颜色c: 2.询问节点a到节点b路径上的 ...
- BZOJ2243 (树链剖分+线段树)
Problem 染色(BZOJ2243) 题目大意 给定一颗树,每个节点上有一种颜色. 要求支持两种操作: 操作1:将a->b上所有点染成一种颜色. 操作2:询问a->b上的颜色段数量. ...
- POJ3237 (树链剖分+线段树)
Problem Tree (POJ3237) 题目大意 给定一颗树,有边权. 要求支持三种操作: 操作一:更改某条边的权值. 操作二:将某条路径上的边权取反. 操作三:询问某条路径上的最大权值. 解题 ...
- bzoj4034 (树链剖分+线段树)
Problem T2 (bzoj4034 HAOI2015) 题目大意 给定一颗树,1为根节点,要求支持三种操作. 操作 1 :把某个节点 x 的点权增加 a . 操作 2 :把某个节点 x 为根的子 ...
- HDU4897 (树链剖分+线段树)
Problem Little Devil I (HDU4897) 题目大意 给定一棵树,每条边的颜色为黑或白,起始时均为白. 支持3种操作: 操作1:将a->b的路径中的所有边的颜色翻转. 操作 ...
- Aizu 2450 Do use segment tree 树链剖分+线段树
Do use segment tree Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://www.bnuoj.com/v3/problem_show ...
- 【POJ3237】Tree(树链剖分+线段树)
Description You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edg ...
随机推荐
- UVA 111 简单DP 但是有坑
题目传送门:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=18201 其实是一道不算难的DP,但是搞了好久,才发现原来是题目没 ...
- HDU 2035 人见人爱A^B(二分求幂,快速求幂)
题意:求A的B次方的后三位数字 思路1:常规求幂,直接取余求解 代码: #include<iostream> #include<cstdio> using namespace ...
- linkedin databus介绍——监听数据库变化,有新数据到来时通知其他消费者app,新数据存在内存里,多份快照
概要结构如下图. 图中显示:Search Index和Read Replicas等系统是Databus的消费者.当主OLTP数据库发生写操作时,连接其上的中继系统会将数据拉到中继中.签入在Search ...
- linux安装与卸载软件
在ubuntu系统中,通常使用apt-get命令完成对软件的安装与卸载 安装的软件通常都放置在一些源中,国内有很多镜像源供下载使用,而系统设置的源保存在目录/etc/apt/sources.list文 ...
- 洛谷 P2919 [USACO08NOV]守护农场Guarding the Farm
题目描述 The farm has many hills upon which Farmer John would like to place guards to ensure the safety ...
- codevs 1144 守望者的逃离
传送门 1144 守望者的逃离 2007年NOIP全国联赛普及组 时间限制: 1 s 空间限制: 128000 KB 题目等级 : 黄金 Gold 题解 题目描述 Description 恶 ...
- 相对路径转绝对路径C++实现
#include<iostream> #include<string> #include<vector> using namespace std; //相对路径转绝 ...
- JS---设置简易红绿灯
<!DOCTYPE html> <html> <head lang="en"> <meta charset="UTF-8&quo ...
- jQuery contextMenu使用
地址:jQuery contextMenu 需要以下文件: jquery.contextMenu.css jquery.min.css jquery.contextMenu.js jquery.ui. ...
- iview组件select之默认展示label,并传空value做方法入参
要求: 默认查询操作日期在当日的数据:(打开页面时默认选中时间.全部) 后台约定:选定“全部”这个条件,传的值是空"" 综上:使用select选择框的v-model绑定数据,使用: ...