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);
      }