A Simple Problem with Integers - POJ 3468 线段树-优快云博客

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/38459123

本文介绍了一种解决区间更新与区间求和问题的方法——线段树。通过具体实例演示了如何构建线段树，并提供了完整的AC代码实现，有助于读者理解和掌握线段树的基本原理及其应用。

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A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 60899		Accepted: 18576
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

题意：区间更新，区间查询的线段树。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
struct node
{ int l,r;
  ll sum,add;
}tree[400010];
void down(int tr)
{ if(tree[tr].add!=0)
  { tree[tr*2].sum+=tree[tr].add*(tree[tr*2].r-tree[tr*2].l+1);
    tree[tr*2+1].sum+=tree[tr].add*(tree[tr*2+1].r-tree[tr*2+1].l+1);
    tree[tr*2].add+=tree[tr].add;
    tree[tr*2+1].add+=tree[tr].add;
    tree[tr].add=0;
  }
}
void build(int tr,int l,int r)
{ tree[tr].l=l;
  tree[tr].r=r;
  if(l==r)
  { scanf("%I64d",&tree[tr].sum);
    tree[tr].add=0;
    return;
  }
  int mid=(l+r)/2;
  build(tr*2,l,mid);
  build(tr*2+1,mid+1,r);
  tree[tr].sum=tree[tr*2].sum+tree[tr*2+1].sum;
  tree[tr].add=0;
}
void update(int tr,int l,int r,ll add)
{ if(tree[tr].l==l && tree[tr].r==r)
  { tree[tr].sum+=(r-l+1)*add;
    tree[tr].add+=add;
    return;
  }
  down(tr);
  int mid=(tree[tr].l+tree[tr].r)/2;
  if(r<=mid)
   update(tr*2,l,r,add);
  else if(l>mid)
   update(tr*2+1,l,r,add);
  else
  { update(tr*2,l,mid,add);
    update(tr*2+1,mid+1,r,add);
  }
  tree[tr].sum=tree[tr*2].sum+tree[tr*2+1].sum;
}
long long query(int tr,int l,int r)
{ if(tree[tr].l==l && tree[tr].r==r)
   return tree[tr].sum;
  down(tr);
  int mid=(tree[tr*2].l+tree[tr*2+1].r)/2;
  if(r<=mid)
   return query(tr*2,l,r);
  else if(l>=mid+1)
   return query(tr*2+1,l,r);
  else
   return query(tr*2,l,mid)+query(tr*2+1,mid+1,r);
}
char s[10];
int main()
{ int n,m,i,j,k,l,r;
  ll add;
  scanf("%d%d",&n,&m);
  build(1,1,n);
  while(m--)
  { scanf("%s",s);
    if(s[0]=='Q')
    { scanf("%d%d",&l,&r);
      printf("%I64d\n",query(1,l,r));
    }
    else
    { scanf("%d%d%I64d",&l,&r,&add);
      update(1,l,r,add);
    }
  }
}