[线段树] POJ 3468 A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 58564 Accepted: 17839Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... ,A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁,A₂, ... ,A_N. -1000000000 ≤A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... ,A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

The sums may exceed the range of 32-bit integers.

 

换个意思表达这个题目，就是说：

给定Q(1 ≤ Q≤ 100,000)个数A1,A2… AQ,，以及可能多次进行的两个操作:
1)对某个区间Ai … Aj的每个数都加n(n可变）2) 求某个区间Ai … Aj的数的和

 

首先要考虑的就是这个树结点里面存放的信息．如果每个节点里面只放区间数的和，那么在执行求区间和操作时速度会很慢，肯定过不了！

可以这样更新：存放一个区间数的和nSum，再存放一个增量Ｃ的累加Inc，　那么求区间和的操作可执行为nSum+Inc*(R-L+1) ．

 

在增加时，如果要加的区间正好覆盖一个节点，则增加其节点的Inc值，不再往下走，否则要更新nSum(加上本次增量),再将增量往下传。这样更新的复杂度就是O(log(n))在查询时，如果待查区间不是正好覆盖一个节点，就将节点的Inc往下带，然后将Inc代表的所有增量累加到nSum上后将Inc清0，接下来再往下查询。Inc往下带的过程也是区间分解的过程，复杂度也是O(log(n))．

 

这回用指针来做，顺便练习一下指针的用法。

 

 

 

 

 

#include <iostream>#include <cstdio>using namespace std;struct CNode{    int L,R;    CNode *pLeft,*pRight;    long long nSum; //原来的和    long long Inc; //增量C的累加};CNode Tree[400010];int nCount=0;int Mid(CNode *pRoot){    return (pRoot->L + pRoot->R)/2;}void BuildTree(CNode *pRoot,int L,int R){    pRoot->L=L;    pRoot->R=R;    pRoot->nSum=0;    pRoot->Inc=0;    if(L==R) return ;    nCount++;    pRoot->pLeft=Tree+nCount;    nCount++;    pRoot->pRight=Tree+nCount;    BuildTree(pRoot->pLeft,L,(L+R)/2);    BuildTree(pRoot->pRight,(L+R)/2+1,R);}void Insert (CNode *pRoot,int i ,int v){    if (pRoot->L == pRoot->R ){        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 += c*(b-a+1);    if(b<=(pRoot->L + pRoot->R)/2)        Add(pRoot->pLeft,a,b,c);    else if(a>=(pRoot->L + pRoot->R)/2+1)        Add(pRoot->pRight,a,b,c);    else{            Add(pRoot->pLeft,a,(pRoot->L+pRoot->R)/2,c);            Add(pRoot->pRight,(pRoot->L+pRoot->R)/2+1,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 +1)*pRoot->Inc;    pRoot->nSum += (pRoot->R - pRoot->L + 1)* pRoot->Inc;    Add(pRoot->pLeft , pRoot->L , Mid(pRoot) , pRoot->Inc );    Add(pRoot->pRight, Mid(pRoot)+1 , pRoot->R , pRoot->Inc);    pRoot->Inc=0;    if( b <= Mid(pRoot) )         return QuerynSum(pRoot->pLeft,a,b);    else if ( a >= Mid(pRoot)+1 )         return QuerynSum(pRoot->pRight,a,b);    else {        return ( QuerynSum(pRoot->pLeft , a , Mid(pRoot) ) +                 QuerynSum(pRoot->pRight, Mid(pRoot)+1, b));    }}int main(){    int n,q,a,b,c;    char cmd[10];    scanf("%d%d",&n,&q);    int i;    nCount=0;    BuildTree(Tree,1,n);    for (i=1 ; i <= n;i++)    {        scanf("%d",&a);        Insert(Tree,i,a);    }    for (i=0;i<q;i++)    {        scanf("%s",cmd);        if(cmd[0]=='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 0;}

关于线段树的题目已经基本建立起框架了，线段树的集训先告一段落，换下一个专题DP啦。