Codeforces Round #254 (Div. 2) E. DZY Loves Colors(线段树 成段更新)

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 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 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.

题目大意：

1 l r x操作 讲 [l，r]上的节点涂成x颜色，并且每个节点的值都加上 |y-x| y为涂之前的颜色

2 l r  操作，求出[l,r]上的和。

思路分析：

当一个区间为相同的颜色。才可以直接涂色，累加sum。

所以我们update时，当区间颜色一致才更新，否则继续访问子节点，也不会损失多少时间

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
using namespace std;
#define root 1,n,1
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long ll;
const int N = 1e5+100;
ll sum[N<<2],col[N<<2],add[N<<2];
void build(int l,int r,int rt)
{
    if(l==r)
    {
        col[rt]=l; return ;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
void pushdown(int rt,int len)
{
    if(col[rt])
    {
        sum[rt<<1]+=add[rt]*(len-(len>>1));
        sum[rt<<1|1]+=add[rt]*(len>>1);
        add[rt<<1]+=add[rt];
        add[rt<<1|1]+=add[rt];
        add[rt]=0;
        col[rt<<1]=col[rt<<1|1]=col[rt];
    }
}
void pushup(int rt)
{
    if(col[rt<<1]==col[rt<<1|1]) col[rt]=col[rt<<1];
    else col[rt]=0;
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void update(int L,int R,int c,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        if(col[rt])
        {
            add[rt]+=abs(c-col[rt]);
            sum[rt]+=abs(c-col[rt])*(r-l+1);
            col[rt]=c;
            return ;
        }
    }
    pushdown(rt,r-l+1);
    int m=(l+r)>>1;
    if(L<=m)update(L,R,c,lson);
    if(R>m) update(L,R,c,rson);
    pushup(rt);
}
ll query(int L,int R,int l,int r,int rt)
{
    ll ans=0;
    if(L<=l&&r<=R)
    {
        return sum[rt];
    }
    pushdown(rt,r-l+1);
    int m=(l+r)>>1;
    if(L<=m) ans+=query(L,R,lson);
    if(R>m) ans+=query(L,R,rson);
    return ans;
}
int main()
{
    int n,m;
    cin>>n>>m;
    build(root);
    for(int i=1;i<=m;i++)
    {
        int op;
        scanf("%d",&op);
        if(op==1)
        {
            int l,r,x;
            scanf("%d%d%d",&l,&r,&x);
            update(l,r,x,root);
        }
        else
        {
            int l,r;
            scanf("%d%d",&l,&r);
            cout<<query(l,r,root)<<endl;
        }
    }
}