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

  1. 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 ...

  2. 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 ...

  3. 【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 ...

  4. 【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 ...

  5. 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. ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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 ...

  10. 【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

  1. 详解Java异常Throwable、Error、Exception、RuntimeException的区别

    在Java中,根据错误性质将运行错误分为两类:错误和异常. 在Java程序的执行过程中,如果出现了异常事件,就会生成一个异常对象.生成的异常对象将传递Java运行时系统,这一异常的产生和提交过程称为抛 ...

  2. BZOJ 1041 [HAOI2008]圆上的整点:数学【费马平方和定理】

    题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1041 题意: 给定n(n <= 2*10^9),问你在圆x^2 + y^2 = n^ ...

  3. JS判断数字、中文、小数位数

    1.JS判断数字 ①var value=$("#test").val(); if(!isNaN(value)){ alert("是数字"); }else{ al ...

  4. spring2.5和struts1.3.8整合

    第一步:导入对应jar文件 第二步: 1.在web容器中实例化spring容器 <!-- 指定spring的配置文件,默认从web根目录寻找配置文件,我们可以通过spring提供的classpa ...

  5. POJ2155 Matrix(二维树状数组||区间修改单点查询)

    Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row an ...

  6. FFmpeg命令:几种常见场景下的FFmpeg命令(摄像头采集推流,桌面屏幕录制推流、转流,拉流等等)

    前提: 首先你得有FFmpeg(ffmpeg官网快捷通道:http://ffmpeg.org/) 再者,推流你得有个流媒体服务,个人测试用小水管:rtmp://eguid.cc:1935/rtmp/t ...

  7. poj 2420 A Star not a Tree?——模拟退火

    题目:http://poj.org/problem?id=2420 精度设成1e-17,做三遍.ans设成double,最后再取整. #include<iostream> #include ...

  8. python3 + selenium + eclipse 中报错:'chromedriver' executable needs to be in PATH. Please see https://sites.google.com/a/chromium.org/chromedriver/home

    解决:提示chrome driver没有放置在正确的路径下,于是下载chrome dirver,然后放置到C:\Python36的目录下,再次运行就OK了!

  9. POJ 1127 Jack Straws (线段相交)

    题意:给定一堆线段,然后有询问,问这两个线段是不是相交,并且如果间接相交也可以. 析:可以用并查集和线段相交来做,也可以用Floyd来做,相交就是一个模板题. 代码如下: #pragma commen ...

  10. win10彻底关闭自动更新

    第1步 https://jingyan.baidu.com/article/9faa7231e7b78b473c28cbb6.html 第2步 http://www.360doc.com/conten ...