DZY Loves Colors
64-bit integer IO format: %I64d Java class name: (Any)
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:
- 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 Input
3 3
1 1 2 4
1 2 3 5
2 1 3
8
3 4
1 1 3 4
2 1 1
2 2 2
2 3 3
3
2
1
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
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
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int maxn = 100010; 5 struct node{ 6 int lt,rt,color; 7 LL sum,len,add; 8 }tree[maxn<<2]; 9 void pushup(int v){ 10 tree[v].sum = tree[v<<1].sum + tree[v<<1|1].sum; 11 if(tree[v<<1].color == tree[v<<1|1].color) 12 tree[v].color = tree[v<<1].color; 13 else tree[v].color = 0; 14 } 15 void pushdown(int v){ 16 if(tree[v].add) { 17 tree[v<<1].add += tree[v].add; 18 tree[v<<1|1].add += tree[v].add; 19 tree[v<<1].sum += tree[v].add*tree[v<<1].len; 20 tree[v<<1|1].sum += tree[v].add*tree[v<<1|1].len; 21 tree[v].add = 0; 22 } 23 if(tree[v].color){ 24 tree[v<<1].color = tree[v<<1|1].color = tree[v].color; 25 tree[v].color = 0; 26 } 27 } 28 void build(int lt,int rt,int v){ 29 tree[v].lt = lt; 30 tree[v].rt = rt; 31 tree[v].len = rt - lt + 1; 32 tree[v].add = 0; 33 tree[v].sum = 0; 34 if(lt == rt){ 35 tree[v].color = lt; 36 return; 37 } 38 int mid = (lt + rt)>>1; 39 build(lt,mid,v<<1); 40 build(mid+1,rt,v<<1|1); 41 pushup(v); 42 } 43 void update(int lt,int rt,int color,int v){ 44 if(lt <= tree[v].lt && rt >= tree[v].rt && tree[v].color){ 45 tree[v].add += abs(tree[v].color - color); 46 tree[v].sum += abs(tree[v].color - color)*tree[v].len; 47 tree[v].color = color; 48 return; 49 } 50 pushdown(v); 51 if(lt <= tree[v<<1].rt) update(lt,rt,color,v<<1); 52 if(rt >= tree[v<<1|1].lt) update(lt,rt,color,v<<1|1); 53 pushup(v); 54 } 55 LL query(int lt,int rt,int v){ 56 if(lt <= tree[v].lt && rt >= tree[v].rt) return tree[v].sum; 57 pushdown(v); 58 LL sum = 0; 59 if(lt <= tree[v<<1].rt) sum += query(lt,rt,v<<1); 60 if(rt >= tree[v<<1|1].lt) sum += query(lt,rt,v<<1|1); 61 pushup(v); 62 return sum; 63 } 64 int main(){ 65 int n,m,op,x,y,color; 66 scanf("%d%d",&n,&m); 67 build(1,n,1); 68 while(m--){ 69 scanf("%d%d%d",&op,&x,&y); 70 if(op == 1){ 71 scanf("%d",&color); 72 update(x,y,color,1); 73 }else printf("%I64d ",query(x,y,1)); 74 } 75 return 0; 76 }