POJ-3468 A Simple Problem with Integers（线段树段变化+段查询模板）

Posted 2020-10-02 GoesOn

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 116441		Accepted: 36178
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

4
55
9
15

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

 

题目大意：给N个数，对应编号1~N。对应两种操作：①求编号a-b对应数字的和。②编号a-b对应数字都加c。

 

解题思路：线段树模板，用lazy思想，关键是要弄懂pushdown函数。理解了的话这类题目都可以看成是模板题。。
个人对pushdown函数的理解：先用lazy[num]存放的是这个区间上所有的数应该变换的值，即编号为num的结点上的数实际值应该是当前值加上lazy[num]。于是对应左右结点对应的区间的和应该加上lazy[num]*对应长度。然后将lazy传给左右结点。当要使用到时再调用pushdown。（这个看个人理解。。这种想法可能只适合我自己。。。做题的时候弄懂的）

 

#include <cstdio>
#include <algorithm>
using namespace std;
const int N=100005;
const int maxn = N*3;
long long a[maxn],val[maxn],lazy[maxn];
int n,m,ql,qr,A,B,value;

void pushdown(int num,int l)
{
    if(lazy[num])
    {
        lazy[num*2] += lazy[num];
        lazy[num*2+1] += lazy[num];

        val[num*2] += (long long)lazy[num]*(l-(l/2));
        val[num*2+1] += (long long)lazy[num]*(l/2);

        lazy[num] = 0;
    }
}

void build(int num,int l,int r)
{
    if(l==r)
    {
        val[num] = a[l];
        return ;
    }
    int mid = (l+r)/2;
    build(num*2,l,mid);
    build(num*2+1,mid+1,r);
    val[num] = val[num*2]+val[num*2+1];
}

long long Findsum(int num,int l,int r)
{
    int mid = (l+r)/2;
    long long ans = 0;
    if(ql<=l&&qr>=r)
    {
        return val[num];
    }
    else
    {
        pushdown(num,r-l+1);
        if(mid>=ql)
        {
            ans += Findsum(num*2,l,mid);
        }
        if(mid<qr)
        {
            ans += Findsum(num*2+1,mid+1,r);
        }
    }
    return ans;
}

void update(int num,int l,int r)
{
    if(A<=l&&B>=r)
    {
        lazy[num] += (long long)value;
        val[num] += (long long)value*(r-l+1);
        return ;
    }
    pushdown(num,r-l+1);
    int mid = (l+r)/2;
    if(A<=mid)
        update(num*2,l,mid);
    if(B>mid)
        update(num*2+1,mid+1,r);
    val[num] = val[num*2]+val[num*2+1];
}

int main()
{
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
    }
    build(1,1,n);
    char c;
    for(int i=1;i<=m;i++)
    {
        getchar();
        scanf("%c",&c);
        if(c==‘Q‘)
        {
            scanf("%d %d",&ql,&qr);
            printf("%lld\n",Findsum(1,1,n));
        }
        else
        {
            scanf("%d %d %d",&A,&B,&value);
            update(1,1,n);
        }
    }
}