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 31 1 2 41 2 3 52 1 3

output

8

input

3 41 1 3 42 1 12 2 22 3 3

output

321

input

10 61 1 5 31 2 7 91 10 10 111 3 8 121 1 10 32 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|1typedef 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;        }    }}