Poj3468 A Simple Problem with Integers 线段树、区间更新-优快云博客

本文链接：https://blog.youkuaiyun.com/Megumin/article/details/51989058

本文介绍了一种使用线段树解决区间加法更新和区间求和查询的问题，通过在节点中增加一个add变量来记录区间内所有点的增量值，实现O(logn)的时间复杂度。

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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

经典的线段树题，在节点中增加一个add变量，记录该区间内所有点都增加的值，即可做到在O( log n ) 内做到区间更新。

#include <iostream>
#include <cstdio>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
#define INF 0x3f3f3f3f
#define LL __int64
LL v;
int a,b;
struct node{
    int l,r;
    LL sum,add;
    int mid(){
        return (l+r)/2;
    }
};
node tree[400005];
void buildtree(int root,int l,int r)
{
    tree[root].l=l;
    tree[root].r=r;
    tree[root].sum=tree[root].add=0;
    if(l!=r){
        buildtree(2*root+1,l,(l+r)/2);
        buildtree(2*root+2,(l+r)/2+1,r);
    }
}
void vinsert(int root,int i,LL v)
{
    if(tree[root].l==tree[root].r){
        tree[root].sum=v;
        return;
    }
    tree[root].sum+=v;
    if(i<=tree[root].mid()){
        vinsert(2*root+1,i,v);
    }else{
        vinsert(2*root+2,i,v);
    }
}
void update(int root,int a,int b,LL v)
{
    if(tree[root].l==a&&tree[root].r==b){
        tree[root].add+=v;
        return;
    }
    tree[root].sum+=v*(b-a+1);
    if(b<=tree[root].mid()){
        update(2*root+1,a,b,v);
    }else if(a>tree[root].mid()){
        update(2*root+2,a,b,v);
    }else{
        update(2*root+1,a,tree[root].mid(),v);
        update(2*root+2,tree[root].mid()+1,b,v);
    }
}
LL answersearch(int root,int a,int b)
{
    LL answer=0;
    if(tree[root].l==a&&tree[root].r==b){
        return tree[root].sum+tree[root].add*(b-a+1);
    }
    tree[root].sum+=tree[root].add*(tree[root].r-tree[root].l+1);
    update(2*root+1,tree[root].l,tree[root].mid(),tree[root].add);
    update(2*root+2,tree[root].mid()+1,tree[root].r,tree[root].add);
    tree[root].add=0;
    if(b<=tree[root].mid()){
        return answersearch(2*root+1,a,b);
    }else if(a>tree[root].mid()){
        return answersearch(2*root+2,a,b);
    }else{
        return answersearch(2*root+1,a,tree[root].mid())+
        answersearch(2*root+2,tree[root].mid()+1,b);
    }
    return answer;
}
int main()
{
    int n,m;
    char ch;
    while(~scanf("%d%d",&n,&m)){
        buildtree(0,1,n);
        for(int i=1;i<=n;i++){
            scanf("%I64d",&v);
            vinsert(0,i,v);
        }
        while(m--){
            getchar();
            scanf("%c",&ch);
            if(ch=='C'){
                scanf("%d%d%I64d",&a,&b,&v);
                update(0,a,b,v);
            }else{
                scanf("%d%d",&a,&b);
                printf("%I64d\n",answersearch(0,a,b));
            }
        }
    }
    return 0;
}