POJ - 3468 A Simple Problem with Integers （线段树——懒标记）（注释详解）

POJ - 3468 A Simple Problem with Integers （模板题）（懒标记）

 

Problem

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.

 

---

---

题意：

给你n个整数和q次询问，你需要区间求和和区间修改

Q 表示询问区间a到b的和；C 表示对区间a到b的每个值加一

思路：

模板题

代码：

#include
#include
#include
using namespace std;
const long long maxn=100005;

long long a[maxn];
struct Node
{
    long long l,r,data;
    long long len,lazy;//lazy懒标记，len记录区间长度，方便求和
} node[maxn*4];
void build(long long l, long long r, long long root) //创建线段树
{
    node[root].l=l, node[root].r=r;
    node[root].len=r-l+1;
    node[root].lazy=0;
    if(l==r)//叶子结点，直接赋值
    {
        node[root].data=a[l];
        return;
    }
    long long mid=(l+r)/2;
    build(l,mid,root<<1);//递归构造左儿子结点
    build(mid+1,r,root<<1|1);//递归构造右儿子结点
    node[root].data = node[root<<1].data + node[root<<1|1].data;//更新父结点
}
void pushDown(long long root)//更新懒标记
{
    if(node[root].lazy)
    {
        node[root*2].lazy += node[root].lazy;
        node[root*2+1].lazy += node[root].lazy;
        node[root*2].data += node[root*2].len * node[root].lazy;
        node[root*2+1].data += node[root*2+1].len * node[root].lazy;
        node[root].lazy=0;
    }
}
void rangeUpdate(long long l,long long r,long long val,long long root)//区间更新
{
    long long L=node[root].l;
    long long R=node[root].r;
    if(l<=L&&R<=r)//当前结点的区间真包含于待查询区间，更新结点信息
    {
        node[root].lazy+=val;
        node[root].data+=node[root].len*val;
    }
    else
    {
        pushDown(root);//关键，查询lazy标记，更新子树
        long long m=(L+R)/2;
        if(l<=m)
            rangeUpdate(l,r,val,root<<1);//左子树和要更新区间有交集
        if(r>m)
            rangeUpdate(l,r,val,root<<1|1);//右子树和要更新区间有交集
        node[root].data=node[root<<1].data+node[root<<1|1].data;
    }
}
long long rangeQueries(long long l,long long r,long long root)//区间查询
{
    long long L=node[root].l;
    long long R=node[root].r;
    if(l<=L&&R<=r)
    {
        return node[root].data;//当前结点的区间真包含于待查询区间，返回结点信息
    }
    else
    {
        pushDown(root);//查询lazy标记，更新子树
        long long m=(L+R)/2;
        if(r<=m)
            return rangeQueries(l,r,root<<1);//要更新区间在左子树
        else if(l>m)
            return rangeQueries(l,r,root<<1|1);//要更新区间在右子树
        else
            return rangeQueries(l,r,root<<1)+rangeQueries(l,r,root<<1|1);//要更新区间在左右子树
    }
}
int main()
{
    int N,Q;
    scanf("%d%d",&N,&Q);
    for(int i=1; i<=N; i++)
        scanf("%lld",&a[i]);
    build(1,N,1);
    for(int i=1; i<=Q; i++)
    {
        char ch;
        long long a,b,c;
        scanf(" %c",&ch);
        if(ch=='Q')
        {
            scanf("%lld%lld",&a,&b);
            printf("%lld\n",rangeQueries(a,b,1));
        }
        else if(ch=='C')
        {
            scanf("%lld%lld%lld",&a,&b,&c);
            rangeUpdate(a,b,c,1);
        }
    }
    return 0;
}

推荐一个：

线段树入门详解