poj4047(线段树+延迟更新)

//线段树延迟更新

//将连续相邻的k个数据的和看成一个元素，建立线段树；交换、替代两种操作都可以统一成加法操作

 

//-------加法

//若当前加法操作的区间和当前所在节点的左右区间相同，则只修改延迟标记adi，和当前区间的最大值sum；若不一致，才开始向下执行加法操作，同时修改最大值

 

//------查询

//若当前查询的区间和当前所在区间相同，则直接返回结果；若不一致，则向下执行加法操作，同时修改最大值

 

 

#include <cstdio>
#include <cstring>
#include <algorithm>
#include<iostream>
using namespace std;
#define MAX 200009
typedef  int ll;

int min(int a,int b)
{
 return a<b?a:b;
}

int max(int a,int b)
{
 return a<b?b:a;
}

ll a[MAX],da[MAX],ans;
struct node 
{
 ll left,right,add;
 ll sum;
}tree[MAX*5];

//*****************************************************
//建立以left,right为左右边界，将数组da中元素存储在首地址从1开始的tree数组中

int build( ll id, ll left, ll right )
{
 tree[id].add=0;
 tree[id].left = left;
 tree[id].right = right;
 if( left == right )
 {
  tree[id].sum = da[left];
  return tree[id].sum;
 }
 else
 {
  ll mid = ( left + right )>>1;   
  tree[id].sum=max(build( id <<1, left, mid ), build( id<<1|1, mid + 1, right ));
  return tree[id].sum;
 }
}
//*****************************************************************

void down(int id)
{
 if(tree[id].left==tree[id].right)
  return ;
 tree[id<<1].add+=tree[id].add;
 tree[id<<1|1].add+=tree[id].add;
 tree[id<<1].sum+=tree[id].add;
 tree[id<<1|1].sum+=tree[id].add;
 tree[id].add=0;    
}

//修改
//****************************************************
//对区间[left,right]内每个元素进行加adi操作
void updata( ll id, ll left, ll right, ll adi)
{
 if(tree[id].left==left&&tree[id].right==right)
 {
  tree[id].add+=adi;
  tree[id].sum+=adi;
  return ;
 }
 else 
 {
  if(tree[id].add!=0)
  {
   down(id);
  }
  
  ll mid=(tree[id].left+tree[id].right)>>1;
  if(right<=mid)
   updata(id<<1,left,right,adi);
  else if(left>mid)
   updata(id<<1|1,left,right,adi);
  else
  {
   updata(id<<1,left,mid,adi);
   updata(id<<1|1,mid+1,right,adi);
  }
  if(tree[id].left!=tree[id].right)
   tree[id].sum=max(tree[id<<1].sum ,tree[id<<1|1].sum);
 }
}
//*****************************************************************


//3.查询
//*****************************************************
//查询区间[left,right]的和
int query(ll id, ll left, ll right)
{
 if( tree[id].left==left&&tree[id].right== right)
 {
  return tree[id].sum;
 }
 else
 {
  if(tree[id].add!=0)
  {
   down(id);
  }
  ll mid = (tree[id].left+tree[id].right)>>1;
  if(right<= mid )
   return query( id <<1, left,right);
  else if(left>mid  )
   return query( id <<1|1,left, right);
  return max( query( id <<1, left,mid) ,query( id<<1|1, mid+1,right ));
 }
}
//*****************************************************************

int main()
{
 int n,m,k,i,j,t;
 cin>>t;
 while(t--)
 {
  scanf("%d%d%d",&n,&m,&k);
  for(i=1;i<=n;i++)
  {
   scanf("%d",&a[i]);
   da[i]=0;
  }
  
  for(i=1;i<=k;i++)
   da[1]+=a[i];  
  for(i=2;i<=n-k+1;i++)
   da[i]=da[i-1]-a[i-1]+a[i+k-1];  
  
  build(1,1,n-k+1);
  for(i=0;i<m;i++)
  {
   int op,x,y;
   scanf("%d%d%d",&op,&x,&y);
   if(0==op)//x处用y替换
   {
    updata(1,max(1,x-k+1),min(n-k+1,x),y-a[x]);
    a[x]=y;
   }
   else if(1==op)//将x处和y处元素交换
   {
        if(x==y)
    continue;
    updata(1,max(1,x-k+1),min(n-k+1,x),a[y]-a[x]);
    updata(1,max(1,y-k+1),min(n-k+1,y),a[x]-a[y]);
    int temp=a[x];
    a[x]=a[y];
    a[y]=temp;
   }
   else if(2==op)//输出[x,y]区间中
   {
    ans=query(1,x,y-k+1);//min(y-k+1,n-k+1));
    printf("%d\n",ans);
   }
  }
 }
 return 0;
}