CodeForces 445E DZY Loves Colors

DZY Loves Colors

Time Limit: 2000ms

Memory Limit: 262144KB

This problem will be judged on CodeForces. Original ID: 445E
64-bit integer IO format: %I64d      Java class name: (Any)

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 the i-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 Input

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

Hint

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.

 

Source

Codeforces Round #254 (Div. 2)

 

解题：线段树

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int maxn = ;
 struct node{
     int lt,rt,color;
     LL sum,len,add;
 }tree[maxn<<];
 void pushup(int v){
     tree[v].sum = tree[v<<].sum + tree[v<<|].sum;
     if(tree[v<<].color == tree[v<<|].color)
         tree[v].color = tree[v<<].color;
     else tree[v].color = ;
 }
 void pushdown(int v){
     if(tree[v].add) {
         tree[v<<].add += tree[v].add;
         tree[v<<|].add += tree[v].add;
         tree[v<<].sum += tree[v].add*tree[v<<].len;
         tree[v<<|].sum += tree[v].add*tree[v<<|].len;
         tree[v].add = ;
     }
     if(tree[v].color){
         tree[v<<].color = tree[v<<|].color = tree[v].color;
         tree[v].color = ;
     }
 }
 void build(int lt,int rt,int v){
     tree[v].lt = lt;
     tree[v].rt = rt;
     tree[v].len = rt - lt + ;
     tree[v].add = ;
     tree[v].sum = ;
     if(lt == rt){
         tree[v].color = lt;
         return;
     }
     int mid = (lt + rt)>>;
     build(lt,mid,v<<);
     build(mid+,rt,v<<|);
     pushup(v);
 }
 void update(int lt,int rt,int color,int v){
     if(lt <= tree[v].lt && rt >= tree[v].rt && tree[v].color){
         tree[v].add += abs(tree[v].color - color);
         tree[v].sum += abs(tree[v].color - color)*tree[v].len;
         tree[v].color = color;
         return;
     }
     pushdown(v);
     if(lt <= tree[v<<].rt) update(lt,rt,color,v<<);
     if(rt >= tree[v<<|].lt) update(lt,rt,color,v<<|);
     pushup(v);
 }
 LL query(int lt,int rt,int v){
     if(lt <= tree[v].lt && rt >= tree[v].rt) return tree[v].sum;
     pushdown(v);
     LL sum = ;
     if(lt <= tree[v<<].rt) sum += query(lt,rt,v<<);
     if(rt >= tree[v<<|].lt) sum += query(lt,rt,v<<|);
     pushup(v);
     return sum;
 }
 int main(){
     int n,m,op,x,y,color;
     scanf("%d%d",&n,&m);
     build(,n,);
     while(m--){
         scanf("%d%d%d",&op,&x,&y);
         if(op == ){
             scanf("%d",&color);
             update(x,y,color,);
         }else printf("%I64d\n",query(x,y,));
     }
     return ;
 }