Codeforces Round #254 (Div. 2)E（线段树懒标记）

E. DZY Loves Colors

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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:

1. Paint all the units with numbers between l and r (both inclusive) with color x.
2. 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

题意：两种操作，一种是将一个区间染成一种颜色，然后每个位置的值增加现在颜色与原来颜色差的绝对值

            第二种操是求区间和

            线段树直接搞就可以了，注意懒操作的变量记录的是该区间单个数增加的值，以便累加

            更新的时候直接暴力递归到满足颜色连续相等的区间即可

#include 
   
   
    
    
#include 
    
    
     
     
#include 
     
     
      
      
#include 
      
      
       
       
using namespace std;
#define lson l,m,o<<1
#define rson m+1,r,o<<1|1

typedef long long ll;
const int MAXN = 100010;
int n,m;
ll sum[MAXN<<2];
int col[MAXN<<2];
ll lazy[MAXN<<2];

int sub(int x,int y)
{
    x-=y;
    return x<0 ? -x : x;
}
void pushup(int o)
{
    sum[o]=sum[o<<1]+sum[o<<1|1];
    if(col[o<<1]==col[o<<1|1])col[o]=col[o<<1];
    else col[o]=0;
}
void pushdown(int o,int len)
{
    if(lazy[o])
    {
        int l=(len +1)/2;
        col[o<<1]=col[o<<1|1]=col[o];
        lazy[o<<1]+=lazy[o];
        lazy[o<<1|1]+=lazy[o];
        sum[o<<1]+=lazy[o]*l;
        sum[o<<1|1]+=lazy[o]*(len-l);
        col[o]=lazy[o]=0;
    }
}
void build(int l,int r,int o)
{
    sum[o]=0;
    lazy[o]=0;
    if(l==r){
        col[o]=l;
        return;
    }
    int m=l+r>>1;
    build(lson);
    build(rson);
    pushup(o);
}
void update(int l,int r,int o,int L,int R,int v)
{
    if(L<=l&&r<=R && col[o]){
        lazy[o]+=sub(col[o],v);
        sum[o]+=(ll)sub(col[o],v)*(r-l+1);
        col[o]=v;
        return;
    }
    int m=l+r>>1;
    pushdown(o,r-l+1);
    if(L<=m)update(lson,L,R,v);
    if(m
       
       
         >1; pushdown(o,r-l+1); ll ans=0; if(L<=m)ans+=query(lson,L,R); if(m