POJ 3468 A Simple Problem with Integers（线段树区间更新）

最新推荐文章于 2019-05-04 02:44:36 发布

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本文介绍了一道关于整数区间更新与查询的经典算法题目。该问题涉及到数组操作、区间更新和查询等关键技术点，并通过使用线段树进行优化，以高效解决大量区间更新和查询操作。

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A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 129619		Accepted: 40216
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

题意：给你N个数，M次操作。有两种操作：第一种“C a b c”是把区间[a,b]中的数都加上c，第二种“Q a b”查询区间[a,b]之间数的和。

区间更新模板题，因为操作种类是字符Q或者C，注意字符读入问题，回车要getchar不然会出问题。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=100005;
typedef long long ll;
int n,m,a,b,c,s[maxn];
ll add[4*maxn];
struct node
{
    int l,r;
    ll num;
}seg[maxn<<2];
void pushup(int root)
{
    seg[root].num=seg[root<<1].num+seg[root<<1|1].num;
}
void pushdown(int root,int ln,int rn)
{
    if(add[root])
    {
        add[root<<1]+=add[root];
        add[root<<1|1]+=add[root];
        seg[root<<1].num+=add[root]*ln;
        seg[root<<1|1].num+=add[root]*rn;
        add[root]=0;
    }
}
void build(int root,int l,int r)
{
    add[root]=0;
    seg[root].l=l;
    seg[root].r=r;
    if(l==r)
    {
        seg[root].num=s[l];
        return;
    }
    int mid=(l+r)>>1;
    build(root<<1,l,mid);
    build(root<<1|1,mid+1,r);
    pushup(root);
}
void update(int root,int l,int r,int c)
{
    if(seg[root].l>=l&&seg[root].r<=r)
    {
        seg[root].num+=(seg[root].r-seg[root].l+1)*c;
        add[root]+=c;
        return;
    }
    int mid=(seg[root].l+seg[root].r)>>1;
    pushdown(root,mid-seg[root].l+1,seg[root].r-mid);
    if(l<=mid) update(root<<1,l,r,c);
    if(r>mid) update(root<<1|1,l,r,c);
    pushup(root);
}
ll query(int root,int l,int r)
{
    if(seg[root].l>=l&&seg[root].r<=r)
        return seg[root].num;
    int mid=(seg[root].l+seg[root].r)>>1;
    pushdown(root,mid-seg[root].l+1,seg[root].r-mid);
    ll ans=0;
    if(l<=mid) ans+=query(root<<1,l,r);
    if(r>mid) ans+=query(root<<1|1,l,r);
    return ans;
}
int main()
{
    char t;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&s[i]);
        build(1,1,n);
        getchar();
        while(m--)
        {
            scanf("%c",&t);
            if(t=='Q')
            {
                scanf("%d %d",&a,&b);
                getchar();
                printf("%lld\n",query(1,a,b));
            }
            else
            {
                scanf("%d %d %d",&a,&b,&c);
                getchar();
                update(1,a,b,c);
            }
        }
    }
    return 0;
}