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:

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

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 = 100010;
struct node{
int lt,rt,color;
LL sum,len,add;
}tree[maxn<<2];
void pushup(int v){
tree[v].sum = tree[v<<1].sum + tree[v<<1|1].sum;
if(tree[v<<1].color == tree[v<<1|1].color)
tree[v].color = tree[v<<1].color;
else tree[v].color = 0;
}
void pushdown(int v){
if(tree[v].add) {
tree[v<<1].add += tree[v].add;
tree[v<<1|1].add += tree[v].add;
tree[v<<1].sum += tree[v].add*tree[v<<1].len;
tree[v<<1|1].sum += tree[v].add*tree[v<<1|1].len;
tree[v].add = 0;
}
if(tree[v].color){
tree[v<<1].color = tree[v<<1|1].color = tree[v].color;
tree[v].color = 0;
}
}
void build(int lt,int rt,int v){
tree[v].lt = lt;
tree[v].rt = rt;
tree[v].len = rt - lt + 1;
tree[v].add = 0;
tree[v].sum = 0;
if(lt == rt){
tree[v].color = lt;
return;
}
int mid = (lt + rt)>>1;
build(lt,mid,v<<1);
build(mid+1,rt,v<<1|1);
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<<1].rt) update(lt,rt,color,v<<1);
if(rt >= tree[v<<1|1].lt) update(lt,rt,color,v<<1|1);
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 = 0;
if(lt <= tree[v<<1].rt) sum += query(lt,rt,v<<1);
if(rt >= tree[v<<1|1].lt) sum += query(lt,rt,v<<1|1);
pushup(v);
return sum;
}
int main(){
int n,m,op,x,y,color;
scanf("%d%d",&n,&m);
build(1,n,1);
while(m--){
scanf("%d%d%d",&op,&x,&y);
if(op == 1){
scanf("%d",&color);
update(x,y,color,1);
}else printf("%I64d\n",query(x,y,1));
}
return 0;
}

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