Cf 444C DZY Loves Colors（段树）

DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of thei-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

Paint all the units with numbers between l and r (both inclusive) with color x.
Ask the sum of colorfulness of the units between l and r (both inclusive).

Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Sample test(s)

input
3 3
1 1 2 4
1 2 3 5
2 1 3

output
8

input
3 4
1 1 3 4
2 1 1
2 2 2
2 3 3

output
3
2
1

input
10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10

output
129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.

线段树的操作，因为涉及到·类似染色的问题。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
typedef long long LL;
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int maxn=1e5+100;
LL col[maxn<<2],sum[maxn<<2],d[maxn<<2];//col[i]!=0代表区间颜色都为col[i]；d[i]用于lazy操作
int n,m;
void pushup(int rs)
{
    if(col[rs<<1]==col[rs<<1|1])   col[rs]=col[rs<<1];
    else    col[rs]=0;
    sum[rs]=sum[rs<<1]+sum[rs<<1|1];
}
void pushdown(int rs,int m)
{
    if(col[rs])
    {
        col[rs<<1]=col[rs<<1|1]=col[rs];
        d[rs<<1]+=d[rs];d[rs<<1|1]+=d[rs];
        sum[rs<<1]+=(LL)(m-(m>>1))*d[rs];
        sum[rs<<1|1]+=(LL)(m>>1)*d[rs];
        d[rs]=col[rs]=0;
    }
}
void build(int rs,int l,int r)
{
    if(l==r)
    {
        sum[rs]=0;
        col[rs]=l;
        return ;
    }
    col[rs]=d[rs]=0;
    int mid=(l+r)>>1;
    build(rs<<1,l,mid);
    build(rs<<1|1,mid+1,r);
    pushup(rs);
}
void update(int rs,int x,int y,int l,int r,int c)
{
    if(l>=x&&r<=y&&col[rs])
    {
        sum[rs]+=abs(col[rs]-c)*(LL)(r-l+1);
        d[rs]+=abs(col[rs]-c);
        col[rs]=c;
        return ;
    }
    pushdown(rs,r-l+1);
    int mid=(l+r)>>1;
    if(x<=mid)  update(rs<<1,x,y,l,mid,c);
    if(y>mid)   update(rs<<1|1,x,y,mid+1,r,c);
    pushup(rs);
}
LL query(int rs,int x,int y,int l,int r)
{
//    cout<<"2333  "<<<<endl;
    if(l>=x&&y>=r)
        return sum[rs];
//    if(l==r)  return sum[rs];
    int mid=(l+r)>>1;
    pushdown(rs,r-l+1);
    LL res=0;
    if(x<=mid)  res+=query(rs<<1,x,y,l,mid);
    if(y>mid)   res+=query(rs<<1|1,x,y,mid+1,r);
    return res;
}
int main()
{
 //    std::ios::sync_with_stdio(false);
     int l,r,x,op;
     while(~scanf("%d%d",&n,&m))
     {
          build(1,1,n);
          while(m--)
          {
              scanf("%d",&op);
              if(op==1)
              {
                  scanf("%d%d%d",&l,&r,&x);
                  update(1,l,r,1,n,x);
              }
              else
              {
                  scanf("%d%d",&l,&r);
                  printf("%I64d
",query(1,l,r,1,n));
              }
          }
     }
     return 0;
}