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\n",query(1,l,r,1,n));
}
}
}
return 0;
}
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