POJ 3468：A Simple Problem with Integers（線段樹應用之：區間更新，區間查詢）

Description

You have N integers, A1, A2, … , AN. 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, … , AN. -1000000000 ≤ Ai ≤ 1000000000.

Each of the next Q lines represents an operation.

“C a b c” means adding c to each of Aa, Aa+1, … , Ab. -10000 ≤ c ≤ 10000.

“Q a b” means querying the sum of Aa, Aa+1, … , Ab.

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

4

55

9

15

Hint

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

題解：線段樹應用之區間更新，區間查詢

做了幾道簡單的點修改，區間查詢的線段樹題後，開始嘗試線段樹比較難的區間更新，如果一直更新的葉子結點的話，其時間複雜度也就會變得不那麼樂觀，線段樹這一資料結構優點也就沒那麼明顯了，于是采用lazy标記

剛開始還不太會，參考的别人的代碼：https://blog.csdn.net/qq_40604099/article/details/82594966

代碼如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
const int maxn=100010;

/**結點結構*/
struct Node
{
    int L,R;   //左右端點
    ll v;    //值
    ll len;        //區間長度
    int lazy;      //lazy标記
};
Node tree[4*maxn];      //用數組表示線段樹

ll a[maxn];     //要儲存的資料

/**建立線段樹，并附初值*/
void build_Tree(int root,int L,int R)
{
    tree[root].L=L;
    tree[root].R=R;
    tree[root].len=R-L+1;
    tree[root].lazy=0;
    if(L==R)
    {
        tree[root].v=a[L];
        return ;
    }
    int mid=(tree[root].L+tree[root].R)/2;

    build_Tree(2*root,L,mid);
    build_Tree(2*root+1,mid+1,R);

    tree[root].v=tree[2*root].v+tree[2*root+1].v;
}

/**向下傳遞lazy标記*/
void pushdown(int root)
{
    if(tree[root].lazy)
    {
        tree[root*2].lazy+=tree[root].lazy;
        tree[2*root+1].lazy+=tree[root].lazy;
        tree[2*root].v+=tree[2*root].len*tree[root].lazy;
        tree[2*root+1].v+=tree[2*root+1].len*tree[root].lazy;
        tree[root].lazy=0;
    }
}

/**區間更新*/
void update(int root,int s,int e,int c)
{
    if(tree[root].L>=s&&tree[root].R<=e)
    {
        tree[root].v+=c*tree[root].len;
        tree[root].lazy+=c;
        return ;
    }
    if(tree[root].L>e||tree[root].R<s)
        return;
    if(tree[root].lazy)
        pushdown(root);
    update(2*root,s,e,c);
    update(2*root+1,s,e,c);
    tree[root].v=tree[2*root].v+tree[2*root+1].v;
}

/**查詢*/
ll  Query(int root,int s,int e)
{
    if(tree[root].L>=s&&tree[root].R<=e)
    {
        return tree[root].v;
    }
    if(tree[root].L>e||tree[root].R<s)
        return 0;
    if(tree[root].lazy)
        pushdown(root);
    return Query(2*root,s,e)+Query(2*root+1,s,e);
}
int main()
{
    int n,q;     //表示數的個數以及詢問次數
    scanf("%d%d",&n,&q);
    for(int i=1; i<=n; i++)
        scanf("%lld",&a[i]);
    build_Tree(1,1,n);

    char ch;   //表示是詢問還是修改操作
    int a,b,c;
    for(int i=1; i<=q; i++)
    {
        getchar();
        scanf("%c",&ch);
        if(ch=='Q')
        {
            scanf("%d%d",&a,&b);
            printf("%lld\n",Query(1,a,b));
        }
        else
        {
            scanf("%d%d%d",&a,&b,&c);
            update(1,a,b,c);
        }
    }
    return 0;
}