线段树的指针表示法。

代码还有待消化。。

代码里面多次用到了函数递归,感觉这次对递归又有了深一层的理解。

 #define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std; struct CNode
{
int L, R;
CNode *pLeft, *pRight;
long long nSum;
long long Inc;
};
CNode Tree[ + ]; //2倍叶子节点就够
int nCount = ;
int Mid(CNode *pRoot)
{
return (pRoot->L + pRoot->R) / ;
} void BuildTree(CNode *pRoot, int L, int R)
{
pRoot->L = L;
pRoot->R = R;
pRoot->nSum = ;
pRoot->Inc = ;
if(L == R)
return;
++nCount;
pRoot->pLeft = Tree + nCount;
++nCount;
pRoot->pRight = Tree + nCount;
BuildTree(pRoot->pLeft, L, (L+R)/);
BuildTree(pRoot->pRight, (L+R)/ + , R);
} void Insert(CNode *pRoot, int i, int v)
{
if(pRoot->L == i && pRoot->R == i)
{
pRoot->nSum = v;
return;
}
pRoot->nSum += v;
if(i <= Mid(pRoot))
Insert(pRoot->pLeft, i, v);
else
Insert(pRoot->pRight, i ,v);
} void Add(CNode *pRoot, int a, int b, long long c)
{
if(pRoot->L == a && pRoot->R == b)
{
pRoot->Inc += c;
return;
}
pRoot->nSum += (b - a + ) * c;
if(b <= (pRoot->L + pRoot->R)/)
Add(pRoot->pLeft, a, b, c);
else if(a >= (pRoot->L + pRoot->R)/ + )
Add(pRoot->pRight, a, b, c);
else
{
Add(pRoot->pLeft, a, (pRoot->L + pRoot->R)/, c);
Add(pRoot->pRight, (pRoot->L + pRoot->R)/ + , b, c);
}
} long long QuerynSum(CNode *pRoot, int a, int b)
{
if(pRoot->L == a && pRoot->R == b)
return pRoot->nSum +
(pRoot->R - pRoot->L + ) * pRoot->Inc; pRoot->nSum += (pRoot->R - pRoot->L + ) * pRoot->Inc;
Add(pRoot->pLeft, pRoot->L, Mid(pRoot), pRoot->Inc);
Add(pRoot->pRight, Mid(pRoot) + , pRoot->R, pRoot->Inc);
pRoot->Inc = ;
if(b <= Mid(pRoot))
return QuerynSum(pRoot->pLeft, a, b);
else if(a >= Mid(pRoot) + )
return QuerynSum(pRoot->pRight, a, b);
else
{
return QuerynSum(pRoot->pLeft, a, Mid(pRoot)) +
QuerynSum(pRoot->pRight, Mid(pRoot) + , b);
}
} int main(void)
{
#ifdef LOCAL
freopen("3468in.txt", "r", stdin);
#endif int n, q, a, b, c;
char cmd[];
scanf("%d%d", &n, &q);
int i, j, k;
nCount = ;
BuildTree(Tree, , n);
for (i = ; i <= n; ++i)
{
scanf("%d", &a);
Insert(Tree, i, a);
}
for(i = ; i < q; ++i)
{
scanf("%s", cmd);
if(cmd[] == 'C')
{
scanf("%d%d%d", &a, &b, &c);
Add(Tree, a, b, c);
}
else
{
scanf("%d%d", &a, &b);
printf("%I64d\n", QuerynSum(Tree, a, b));
}
} return ;
}

代码君

POJ 3468 A Simple Problem with Integers的更多相关文章

  1. POJ.3468 A Simple Problem with Integers(线段树 区间更新 区间查询)

    POJ.3468 A Simple Problem with Integers(线段树 区间更新 区间查询) 题意分析 注意一下懒惰标记,数据部分和更新时的数字都要是long long ,别的没什么大 ...

  2. poj 3468 A Simple Problem with Integers 【线段树-成段更新】

    题目:id=3468" target="_blank">poj 3468 A Simple Problem with Integers 题意:给出n个数.两种操作 ...

  3. 线段树(成段更新) POJ 3468 A Simple Problem with Integers

    题目传送门 /* 线段树-成段更新:裸题,成段增减,区间求和 注意:开long long:) */ #include <cstdio> #include <iostream> ...

  4. POJ 3468 A Simple Problem with Integers(分块入门)

    题目链接:http://poj.org/problem?id=3468 A Simple Problem with Integers Time Limit: 5000MS   Memory Limit ...

  5. POJ 3468 A Simple Problem with Integers(线段树功能:区间加减区间求和)

    题目链接:http://poj.org/problem?id=3468 A Simple Problem with Integers Time Limit: 5000MS   Memory Limit ...

  6. poj 3468 A Simple Problem with Integers(线段树+区间更新+区间求和)

    题目链接:id=3468http://">http://poj.org/problem? id=3468 A Simple Problem with Integers Time Lim ...

  7. poj 3468 A Simple Problem with Integers 线段树区间加,区间查询和

    A Simple Problem with Integers Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://poj.org/problem?i ...

  8. poj 3468 A Simple Problem with Integers 线段树区间加,区间查询和(模板)

    A Simple Problem with Integers Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://poj.org/problem?i ...

  9. poj 3468:A Simple Problem with Integers(线段树,区间修改求和)

    A Simple Problem with Integers Time Limit: 5000MS   Memory Limit: 131072K Total Submissions: 58269   ...

  10. poj 3468 A Simple Problem with Integers 线段树第一次 + 讲解

    A Simple Problem with Integers Description You have N integers, A1, A2, ... , AN. You need to deal w ...

随机推荐

  1. js函数延迟执行

    function delay(value){ //全局变量保存当前值 window._myTempDalayValue = value; setTimeout(function(){ //延时之后与全 ...

  2. ASP.NET MVC 从IHttp到页面输出

    MVCHandler应该算是MVC真正开始的地方.MVCHandler实现了IHttpHandler接口,ProcessRequest便是方法入口. MVCHandler : IHttpHandler ...

  3. javascript设计模式--状态模式(State)

    <!doctype html> <html lang="en"> <head> <meta charset="UTF-8&quo ...

  4. Create Script Template In Edit Mode

    很多时候 许多类 的 格式 都是重复的,比如 从配置文件中映射出来的类. 这个时候写一个 类模板 就很节省时间了. Code public static string TestPath = " ...

  5. LVM quick start

    这里记录一些任务用到的快速命令,详细LVM管理可参考: http://wenku.baidu.com/view/c29b8bc4bb4cf7ec4afed0ad.html 1.把home分区的磁盘空间 ...

  6. 【五】PHP数组操作函数

    1.输出数组的结构:bool print_r(数组); $arr=array('jack','mike','tom'); print_r($arr);//Array ( [0] => jack ...

  7. 【面试题】Round A China New Grad Test 2014总结

    我也有够懒的,今天才跑来写总结,自觉面壁中… 上一篇是Practice Round,今天是Round A,五道题. 每次做完都想说,其实题不难..但在做的过程中总是会各种卡,只有自己一行一行实现了,才 ...

  8. HDU 1403 Longest Common Substring(后缀数组,最长公共子串)

    hdu题目 poj题目 参考了 罗穗骞的论文<后缀数组——处理字符串的有力工具> 题意:求两个序列的最长公共子串 思路:后缀数组经典题目之一(模版题) //后缀数组sa:将s的n个后缀从小 ...

  9. MYSQL判断某个表是否已经存在

    方法一.You don't need to count anything. SELECT 1 FROM testtable LIMIT 1; If there's no error, table ex ...

  10. JAVA类型信息——Class对象

    JAVA类型信息——Class对象 一.RTTI概要 1.类型信息RTTI :即对象和类的信息,例如类的名字.继承的基类.实现的接口等. 2.类型信息的作用:程序员可以在程序运行时发现和使用类型信息. ...