线段树 + 区间更新(区间增加v)模板 ---- poj 3468 - Snarl_jsb

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, A 1 , A 2 , ... , 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 A 1 , A 2 , ... , 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 integer
【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；

#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#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);
}
}

int main()
{
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;
}

区间修改模板

Updata(把某一区间的数增加v)

Query（计算区间元素的和，最小值，最大值）

本代码为求的元素的和，求最大值的话改为//的语句

#define MAX 110000
#define LL long long
LL n,m;
LL ans;
struct Tree
{
LL l,r;
LL sum,add;//maxx
};
Tree tree[MAX*4];
void pushup(LL x)
{
LL tmp=2*x;
tree[x].sum=tree[tmp].sum+tree[tmp+1].sum;//tree[x].maxx=max(tree[tmp].maxx,tree[tmp+1].maxx);
}
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].maxx+=tree[x].add;
tree[tmp+1].sum+=tree[x].add*(tree[tmp+1].r-tree[tmp+1].l+1);
//tree[tmp+1].maxx+=tree[x].add;
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);//scanf("%lld",&tree[x].maxx);
return ;
}
int tmp=x<<1;
int mid=(l+r)>>1;
build(l,mid,tmp);
build(mid+1,r,tmp+1);
pushup(x);
}
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);//tree[x].maxx+=c;
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)
return ;
if(l<=tree[x].l&&r>=tree[x].r)
{
ans+=tree[x].sum;//ans=max(ans,tree[x].maxx);
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);
}
}
int main()
{
scanf("%lld %lld",&n,&m);
build(1,n,1);
char ch;
if(ch=='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);
}