POJ 3468 - A Simple Problem with Integers(线段树区间更新)-优快云博客

本文介绍了一种基于线段树的数据结构解决区间更新与查询的问题，适用于处理大量区间加法及区间求和操作，通过递归分解和懒惰传播实现高效算法。

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A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 100889 Accepted: 31452
Case Time Limit: 2000MS
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.

题意：
并查集区间更新查询。

解题思路：
模板。

AC代码：

#include<stdio.h>
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int maxn = 1e5+5;
long long add[maxn<<2];
long long sum[maxn<<2];
void PushUp(int rt) {sum[rt] = sum[rt<<1]+sum[rt<<1|1];}
void PushDown(int rt,int m)
{
    if(add[rt])
    {
        add[rt<<1] += add[rt];
        add[rt<<1|1] += add[rt];
        sum[rt<<1] += add[rt]*(m-(m>>1));
        sum[rt<<1|1] += add[rt]*(m>>1);
        add[rt] = 0;
    }
}
void Update(int L,int R,int c,int l,int r,int rt)
{
    if(L <= l && R >= r)
    {
        add[rt] += c;
        sum[rt] += c*(r-l+1);
        return ;
    }
    PushDown(rt,r-l+1);
    int mid = (l+r)>>1;
    if(L <= mid)  Update(L,R,c,lson);
    if(R > mid)   Update(L,R,c,rson);
    PushUp(rt);
}
long long query(int L,int R,int l,int r,int rt)
{
    if(L <= l && R >= r)    return sum[rt];
    PushDown(rt,r-l+1);
    int mid = (l+r)>>1;
    long long res = 0;
    if(L <= mid)  res += query(L,R,lson);
    if(R > mid)   res += query(L,R,rson);
    return res;
}
void build(int l,int r,int rt)
{
    add[rt] = 0;
    if(l == r)
    {
        scanf("%lld",&sum[rt]);
        return ;
    }
    int mid = (l+r)>>1;
    build(lson);
    build(rson);
    PushUp(rt);
}
int main()
{
    int n,T;
    scanf("%d%d",&n,&T);
    build(1,n,1);
    while(T--)
    {
        char op[2];
        scanf("%s",&op);
        if(op[0] == 'Q')
        {
            int L,R;
            scanf("%d%d",&L,&R);
            printf("%lld\n",query(L,R,1,n,1));
        }
        else
        {
            int L,R,c;
            scanf("%d%d%d",&L,&R,&c);
            Update(L,R,c,1,n,1);
        }
    }
    return 0;
}