POJ 3468 A Simple Problem with Integers(线段树区间操作)

题目的意思很简单就是一个线段树的区间的增加数字，与区间的查询。

说一下，区间操作的题目第一次做啊，我瞎搞了一下，超时了啊。于是求助于啸爷，啸爷又是“苦心教导”啊。。感激不尽啊。。

一个区间当有更新的时候，先把区间上的总和更新一下，然后标记一下更新的多少，然后如果以后还会找到这个区间的时候，要把他所标记的那个数字传到他的左右子树中去。因为，这样的话，只更新了这个区间。他的子区间能没有发生过改变。所以要把他的标记给下面的然后抹去自己的标记。(说明它自己已经完成它对自己子树的更新)。

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 53175		Accepted: 15899
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

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-7
#define M 10001000
//#define LL __LL64
#define LL long long
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
const LL maxn = 100100;

using namespace std;

struct node
{
LL sum;
LL add;
}f[4*maxn];

LL num[4*maxn];

void Bulid(LL l, LL r, LL site)
{
if(l == r)
{
f[site].sum = num[l];
f[site].add = 0;
return;
}
LL mid = (l+r)>>1;
Bulid(l, mid, site<<1);
Bulid(mid+1, r, site<<1|1);
f[site].add = 0;
f[site].sum = f[site<<1].sum+f[site<<1|1].sum;
}

void Updata(LL L, LL R, LL l, LL r, LL site, LL x)
{
if(L == l && R == r)
{
f[site].add += x;
f[site].sum += (r-l+1)*x;
return;
}
LL mid = (L+R)>>1;

if(f[site].add)//这里如果不去掉标记的话，最后的赋值又可能会变小；因为没更新左右的值；
{
Updata(L, mid, L, mid, site<<1, f[site].add);
Updata(mid+1, R, mid+1, R, site<<1|1, f[site].add);
f[site].add = 0;
}

if(r <= mid)
Updata(L, mid, l, r, site<<1, x);
else if(l > mid)
Updata(mid+1, R, l, r, site<<1|1, x);
else
{
Updata(L, mid, l, mid, site<<1, x);
Updata(mid+1, R, mid+1, r, site<<1|1, x);
}
f[site].sum = f[site<<1].sum+f[site<<1|1].sum;
}

node Qusery(LL L, LL R, LL l, LL r, LL site)
{

if(L == l && R == r)
{
return f[site];
}
LL mid = (L + R)>>1;

if(f[site].add)
{
Updata(L, mid, L, mid, site<<1, f[site].add);
Updata(mid+1, R, mid+1, R, site<<1|1, f[site].add);
f[site].add = 0;
}

if(r <= mid)
{
return Qusery(L, mid, l, r, site<<1);
}
else if(l > mid)
{
return Qusery(mid+1, R, l, r, site<<1|1);
}
node t1 = Qusery(L, mid, l, mid, site<<1);
node t2 = Qusery(mid+1, R, mid+1, r, site<<1|1);
t1.sum += t2.sum;
return t1;
}

int main()
{
LL n, m;
while(~scanf("%lld %lld",&n, &m))
{
for(LL i = 1; i <= n; i++)
scanf("%lld",&num[i]);
char str;
LL l, r, x;
Bulid(1, n, 1);
while(m--)
{
scanf("%*c%c",&str);
if(str == 'Q')
{
scanf("%lld %lld",&l, &r);
node temp;
temp = Qusery(1, n, l, r, 1);
printf("%lld\n",temp.sum);
}
else if(str == 'C')
{
scanf("%lld %lld %lld",&l, &r, &x);
Updata(1, n, l, r, 1, x);
}
}
}
return 0;
}