线段树 + 区间更新 + 模板 ---- poj 3468

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 59798		Accepted: 18237
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

【题目大意】

一个数列，每次操作可以是将某区间数字都加上一个相同的整数，也可以是询问一个区间中所有数字的和。（这里区间指的是数列中连续的若干个数）对每次询问给出结果。

【题目分析】

裸的区间更新线段树。

线段树单点更新和区间更新的区别：
1.每个结点中多了一个add值，代表该结点以下的结点需要增加的值；
2.build函数中，如果在建树的过程中就赋值给num，那么在建完树之后不要忘记pushup,因为此时只是叶子结点有值，上面的值都为空；这个在区间更新中很常用，因为区间更新中如果输入一个值，然后更新一个值，这样会很麻烦，会耗费更多的时间；
3.update函数中，区间更新多了一个upshdown函数，并且更新sum和add值的判断条件是树中结点的l~r和要更新的区间的l~r相等，此时sum加的值是整个区间的长度*要更新的值，然后add值记录后面每个结点需要加上的值，即：c;
4.upshdown函数最后不要忘了将延时标记add清零；

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//Memory   Time
//  K      MS
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<iomanip>
#include<string>
#include<climits>
#include<cmath>
#define MAX 110000
#define LL long long
using namespace std;
LL n,m;
LL ans;
struct Tree
{
    LL l,r;
    LL sum,add;
};
Tree tree[MAX*3];
 
void pushup(LL x)
{
    LL tmp=2*x;
    tree[x].sum=tree[tmp].sum+tree[tmp+1].sum;
}
 
 
void pushdown(LL x)
{
    LL tmp=2*x;
    tree[tmp].add+=tree[x].add;
    tree[tmp+1].add+=tree[x].add;
    tree[tmp].sum+=tree[x].add*(tree[tmp].r-tree[tmp].l+1);
    tree[tmp+1].sum+=tree[x].add*(tree[tmp+1].r-tree[tmp+1].l+1);
    tree[x].add=0;
}
 
void build(int l,int r,int x)
{
    tree[x].l=l;
    tree[x].r=r;
    tree[x].add=0;
    if(l==r)
    {
        scanf("%lld",&tree[x].sum);
        return ;
    }
    int tmp=x<<1;
    int mid=(l+r)>>1;
    build(l,mid,tmp);
    build(mid+1,r,tmp+1);
    pushup(x);     //如果在建树的过程中给sum赋值，记得后面要pushup
}
 
 
void update(LL l,LL r,LL c,LL x)
{
    if(r<tree[x].l||l>tree[x].r)
        return ;
    if(l<=tree[x].l&&r>=tree[x].r)
    {
        tree[x].add+=c;
        tree[x].sum+=c*(tree[x].r-tree[x].l+1);
        return ;
    }
    if(tree[x].add)
        pushdown(x);
    LL tmp=x<<1;
    update(l,r,c,tmp);    //  !!!
    update(l,r,c,tmp+1);
    pushup(x);
}
 
 
void query(LL l,LL r,LL x)
{
    if(r<tree[x].l||l>tree[x].r)         //要更新的区间不在该区间上
        return ;
    if(l<=tree[x].l&&r>=tree[x].r)      //要更新区间包括了该区间
    {
        ans+=tree[x].sum;
        return ;
    }
    if(tree[x].add)
        pushdown(x);
    LL tmp=x<<1;
    LL mid=(tree[x].l+tree[x].r)>>1;
    if(r<=mid)
        query(l,r,tmp);
    else if(l>mid)
        query(l,r,tmp+1);
    else
    {
        query(l,mid,tmp);
        query(mid+1,r,tmp+1);
    }
//    pushup(x);
}
 
 
int main()
{
//    freopen("cin.txt","r",stdin);
//    freopen("cout.txt","w",stdout);
    while(~scanf("%lld %lld",&n,&m))
    {
        build(1,n,1);
        char str[5];
        while(m--)
        {
            scanf("%s",str);
            if(str[0]=='Q')
            {
                LL l,r;
                scanf("%lld %lld",&l,&r);
                ans=0;
                query(l,r,1);
                printf("%lld\n",ans);
            }
            else
            {
                LL l,r,c;
                scanf("%lld %lld %lld",&l,&r,&c);
                update(l,r,c,1);
            }
        }
    }
    return 0;
}