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Poj 3468 A Simple Problem with Integers（線段樹區間更新延遲標記）

編輯：C++入門知識

Poj 3468 A Simple Problem with Integers（線段樹區間更新延遲標記）

A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 73239 Accepted: 22607 Case 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 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi
目測自己完全可以手寫了~ 。題目最下面有提示回溯加和的時候小心溢出32位整數。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int INF=100003;
int val[INF];
struct Tree
{
    int left,right;
    long long total,mark;
} tree[INF<<2];

long long  create(int root ,int left,int right)
{
    tree[root].left=left;
    tree[root].right=right;
    tree[root].mark=0;
    if(left==right)
        return  tree[root].total=val[left];
    long long  a,b,mid=(left+right)>>1;
    a=create(root<<1,left,mid);
    b=create(root<<1|1,mid+1,right);
    return tree[root].total=a+b;
}

void update_mark(int root)
{
    if(tree[root].mark)
    {
        tree[root].total+=tree[root].mark*(tree[root].right-tree[root].left+1);
        if(tree[root].left!=tree[root].right)
        {
            tree[root<<1].mark+=tree[root].mark;
            tree[root<<1|1].mark+=tree[root].mark;
        }
        tree[root].mark=0;
    }
}

long long  calculate(int root ,int left,int right)
{
    update_mark(root);
    if(tree[root].left>right||tree[root].right=left&&tree[root].right<=right)
    {
        return tree[root].total;
    }
    long long  a=calculate(root<<1,left,right);
    long long  b=calculate(root<<1|1,left,right);
    return a+b;
}

long long  update(int root , int left,int right,int val)
{
    update_mark(root);
    if(tree[root].left>right||tree[root].right=left&&tree[root].right<=right)
    {
        tree[root].mark+=val;//這裡一點需要注意，應重新更新root結點
        update_mark(root);   //讓其返回當前結點更新後的最新信息。
        return tree[root].total;
    }
    long long   a=update(root<<1,left,right,val);
    long  long b=update(root<<1|1,left,right,val);
    return tree[root].total=a+b;
}

int main()
{
    int n,q;
    while(cin>>n>>q)
    {
        for(int i=1; i<=n; i++)
            scanf(%d,&val[i]);
        create(1,1,n);
        for(int i=0; i