poj3468A Simple Problem with Integers（线段树+成段更新）

最新推荐文章于 2020-07-03 11:31:10 发布

FBI-PC

最新推荐文章于 2020-07-03 11:31:10 发布

阅读量1.4k

点赞数 1

CC 4.0 BY-SA版权

分类专栏：数据结构

本文链接：https://blog.youkuaiyun.com/u014303647/article/details/38947559

数据结构专栏收录该内容

30 篇文章

订阅专栏

本文介绍了一道经典的线段树懒惰传播问题，通过实例详细解析了如何处理区间更新和查询操作。针对大规模数据集，文章提供了高效的C++实现代码。

题目链接：
huangjing
题意：
给n个数，然后有两种操作。
【1】Q a b 询问a到b区间的和。
【2】C a b c将区间a到b的值都增加c。
思路：
线段树成段更新的入门题目。。学会使用lazy即可。还需要注意的是，lazy的时候更改是累加，而不是直接修改。。有可能连续几次进行修改操作。。注意这一点就好了。。。

题目：

Language:

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 62629		Accepted: 19215
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

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=100000+10;
ll col[maxn*4];
ll tree[maxn*4];

int n,m;

void push_up(int dex)
{
    tree[dex]=tree[dex<<1]+tree[dex<<1|1];
}

void push_down(int cnt,int dex)
{
    if(col[dex])
    {
        col[dex<<1]+=col[dex];//这里要累积起来
        col[dex<<1|1]+=col[dex];
        tree[dex<<1]+=(cnt-(cnt>>1))*col[dex];
        tree[dex<<1|1]+=(cnt>>1)*col[dex];
        col[dex]=0;
    }
}

void buildtree(int l,int r,int dex)
{
    col[dex]=0;
    if(l==r)
    {
        scanf("%I64d",&tree[dex]);
        return;
    }
    int mid=(l+r)/2;
    buildtree(l,mid,dex<<1);
    buildtree(mid+1,r,dex<<1|1);
    push_up(dex);
}

void update(int L,int R,int l,int r,int dex,long long val)
{
    if(L<=l&&R>=r)
    {
        tree[dex]+=(r-l+1)*val;
        col[dex]+=val;
        return;
    }
    push_down(r-l+1,dex);
    int mid=(l+r)/2;
    if(L<=mid) update(L,R,l,mid,dex<<1,val);
    if(R>mid)  update(L,R,mid+1,r,dex<<1|1,val);
    push_up(dex);
}

long long Query(int L,int R,int l,int r,int dex)
{
    if(L<=l&&R>=r)
        return tree[dex];
    push_down(r-l+1,dex);
    int mid=(l+r)/2;
    if(R<=mid) return Query(L,R,l,mid,dex<<1);
    else if(L>mid)  return Query(L,R,mid+1,r,dex<<1|1);
    else return Query(L,R,l,mid,dex<<1)+Query(L,R,mid+1,r,dex<<1|1);
}

int main()
{
    char str[2];
    int u,v;
    long long w;
    while(~scanf("%d%d",&n,&m))
    {
        buildtree(1,n,1);
        while(m--)
        {
           scanf("%s",str);
           if(str[0]=='Q')
           {
               scanf("%d%d",&u,&v);
               printf("%I64d\n",Query(u,v,1,n,1));
           }
           else
           {
               scanf("%d%d%I64d",&u,&v,&w);
               update(u,v,1,n,1,w);
           }
        }
    }
    return 0;
}