Aragorn's Story

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1321    Accepted Submission(s): 344

Problem Description
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.
 
Input
Multiple test cases, process to the end of input.

For each case, The first line contains three integers N, M, P which means there will be N(1 ≤ N ≤ 50000) camps, M(M = N-1) edges and P(1 ≤ P ≤ 100000) operations. The number of camps starts from 1.

The next line contains N integers A1, A2, ...AN(0 ≤ Ai ≤ 1000), means at first in camp-i has Ai enemies.

The next M lines contains two integers u and v for each, denotes that there is an edge connects camp-u and camp-v.

The next P lines will start with a capital letter 'I', 'D' or 'Q' for each line.

'I', followed by three integers C1, C2 and K( 0≤K≤1000), which means for camp C1, C2 and all camps on the path from C1 to C2, increase K soldiers to these camps.

'D', followed by three integers C1, C2 and K( 0≤K≤1000), which means for camp C1, C2 and all camps on the path from C1 to C2, decrease K soldiers to these camps.

'Q', followed by one integer C, which is a query and means Aragorn wants to know the number of enemies in camp C at that time.

 
Output
For each query, you need to output the actually number of enemies in the specified camp.
 
Sample Input
3 2 5
1 2 3
2 1
2 3
I 1 3 5
Q 2
D 1 2 2
Q 1
Q 3
 
Sample Output
7
4
8

Hint

1.The number of enemies may be negative.

2.Huge input, be careful.

 
Source
 

裸的树链剖分的入门题;

我是套的树状数组实现的

 /* ***********************************************
Author :kuangbin
Created Time :2013/8/14 23:14:27
File Name :F:\2013ACM练习\专题学习\数链剖分\HDU3966.cpp
************************************************ */
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MAXN = ;
struct Edge
{
int to,next;
}edge[MAXN*];
int head[MAXN],tot;
int top[MAXN];
int fa[MAXN];
int deep[MAXN];
int num[MAXN];
int p[MAXN];
int fp[MAXN];
int son[MAXN];
int pos;
void init()
{
tot = ;
memset(head,-,sizeof(head));
pos = ;//使用树状数组,编号从头1开始
memset(son,-,sizeof(son));
}
void addedge(int u,int v)
{
edge[tot].to = v; edge[tot].next = head[u]; head[u] = tot++;
}
void dfs1(int u,int pre,int d)
{
deep[u] = d;
fa[u] = pre;
num[u] = ;
for(int i = head[u];i != -; i = edge[i].next)
{
int v = edge[i].to;
if(v != pre)
{
dfs1(v,u,d+);
num[u] += num[v];
if(son[u] == - || num[v] > num[son[u]])
son[u] = v;
}
}
}
void getpos(int u,int sp)
{
top[u] = sp;
p[u] = pos++;
fp[p[u]] = u;
if(son[u] == -) return;
getpos(son[u],sp);
for(int i = head[u];i != -;i = edge[i].next)
{
int v = edge[i].to;
if( v != son[u] && v != fa[u])
getpos(v,v);
}
} //树状数组
int lowbit(int x)
{
return x&(-x);
}
int c[MAXN];
int n;
int sum(int i)
{
int s = ;
while(i > )
{
s += c[i];
i -= lowbit(i);
}
return s;
}
void add(int i,int val)
{
while(i <= n)
{
c[i] += val;
i += lowbit(i);
}
}
void Change(int u,int v,int val)//u->v的路径上点的值改变val
{
int f1 = top[u], f2 = top[v];
int tmp = ;
while(f1 != f2)
{
if(deep[f1] < deep[f2])
{
swap(f1,f2);
swap(u,v);
}
add(p[f1],val);
add(p[u]+,-val);
u = fa[f1];
f1 = top[u];
}
if(deep[u] > deep[v]) swap(u,v);
add(p[u],val);
add(p[v]+,-val);
}
int a[MAXN];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int M,P;
while(scanf("%d%d%d",&n,&M,&P) == )
{
int u,v;
int C1,C2,K;
char op[];
init();
for(int i = ;i <= n;i++)
{
scanf("%d",&a[i]);
}
while(M--)
{
scanf("%d%d",&u,&v);
addedge(u,v);
addedge(v,u);
}
dfs1(,,);
getpos(,);
memset(c,,sizeof(c));
for(int i = ;i <= n;i++)
{
add(p[i],a[i]);
add(p[i]+,-a[i]);
}
while(P--)
{
scanf("%s",op);
if(op[] == 'Q')
{
scanf("%d",&u);
printf("%d\n",sum(p[u]));
}
else
{
scanf("%d%d%d",&C1,&C2,&K);
if(op[] == 'D')
K = -K;
Change(C1,C2,K);
}
}
}
return ;
}

HDU 3966 Aragorn's Story (树链剖分+树状数组)的更多相关文章

  1. hdu 3966 Aragorn's Story(树链剖分+树状数组/线段树)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3966 题意: 给出一棵树,并给定各个点权的值,然后有3种操作: I C1 C2 K: 把C1与C2的路 ...

  2. 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 ...

  3. 洛谷 P3384 【模板】树链剖分-树链剖分(点权)(路径节点更新、路径求和、子树节点更新、子树求和)模板-备注结合一下以前写的题目,懒得写很详细的注释

    P3384 [模板]树链剖分 题目描述 如题,已知一棵包含N个结点的树(连通且无环),每个节点上包含一个数值,需要支持以下操作: 操作1: 格式: 1 x y z 表示将树从x到y结点最短路径上所有节 ...

  4. HDU 3966 Aragorn's Story 树链剖分+树状数组 或 树链剖分+线段树

    HDU 3966 Aragorn's Story 先把树剖成链,然后用树状数组维护: 讲真,研究了好久,还是没明白 树状数组这样实现"区间更新+单点查询"的原理... 神奇... ...

  5. hdu 3966 Aragorn&#39;s Story(树链剖分+树状数组)

    pid=3966" target="_blank" style="">题目链接:hdu 3966 Aragorn's Story 题目大意:给定 ...

  6. HDU 3966 Aragorn's Story(模板题)【树链剖分】+【线段树】

    <题目链接> 题目大意: 给定一颗带点权的树,进行两种操作,一是给定树上一段路径,对其上每个点的点权增加或者减少一个数,二是对某个编号点的点权进行查询. 解题分析: 树链剖分的模板题,还不 ...

  7. HDU 3966 /// 树链剖分+树状数组

    题意: http://acm.hdu.edu.cn/showproblem.php?pid=3966 给一棵树,并给定各个点权的值,然后有3种操作: I x y z : 把x到y的路径上的所有点权值加 ...

  8. HDU 5044 (树链剖分+树状数组+点/边改查)

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5044 题目大意:修改链上点,修改链上的边.查询所有点,查询所有边. 解题思路: 2014上海网赛的变 ...

  9. HDU 5293 Train chain Problem - 树链剖分(树状数组) + 线段树+ 树型dp

    传送门 题目大意: 一颗n个点的树,给出m条链,第i条链的权值是\(w_i\),可以选择若干条不相交的链,求最大权值和. 题目分析: 树型dp: dp[u][0]表示不经过u节点,其子树的最优值,dp ...

随机推荐

  1. centos6.5 安装、启动vnc

    一.安装vnc 1.确保当前账号是root2.查看本机是否已经安装vncserver rpm -qa|grep tigervnc 3.安装vncserver yum -y install tigerv ...

  2. 【hdoj_1002】A+B Problem ||(大数)

    题目:http://acm.hdu.edu.cn/showproblem.php?pid=1002 题目提示,相加的两个数的位数可能很大(最多可达1000位),而int最多32位,long long类 ...

  3. webpack2.0构建vue2.0超详细精简版

    原文地址:http://blog.csdn.net/dahuzix/article/details/55549387 npm init -y 初始化项目 安装各种依赖项 npm install --s ...

  4. 错误:Could not create the Android package. See the Output (Build) window for more details

    错误:Could not create the Android package. See the Output (Build) window for more details. Mono For An ...

  5. 基于Redis实现——分布式锁与实现

    实现 使用的是jedis来连接Redis. 实现思想 获取锁的时候,使用setnx加锁,并使用expire命令为锁添加一个超时时间,超过该时间则自动释放锁,锁的value值为一个随机生成的UUID,通 ...

  6. 《java虚拟机》----java内存区域与内存溢出异常

    No1: java虚拟机所管理的内存将会包括以下几个运行时数据区域 1.方法区 2.虚拟机栈 3.本地方法栈 4.堆 5.程序计数器 No2: 程序计数器: 程序计数器(Program Counter ...

  7. shell grep 筛选

    egrep '^.{11}(WP)' inputfile | sort -k1.17,1.24 > outputfile WP 在12位开始.

  8. MAC OS 10.11.1虚拟机免费下载已安装Xcode7图片后有下载地址

    MAC OS 10.11.1虚拟机免费下载已安装Xcode7图片后有下载地址 注意:已经下载过MAC OS 10.10.5虚拟机免费下载(可安装Xcode7)链接:http://www.cnblogs ...

  9. Sea Battle<海战>(思路题)

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  10. 深度理解python中的元类

    本文转自:(英文版)https://stackoverflow.com/questions/100003/what-is-a-metaclass-in-python   (翻译版)   http:// ...