codeforces 444C. DZY Loves Colors （线段树）

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 thei-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:

Can you help DZY?

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.

For each operation 2, print a line containing the answer — sum of colorfulness.

3 31 1 2 41 2 3 52 1 3output
8input
3 41 1 3 42 1 12 2 22 3 3output
321input
10 61 1 5 31 2 7 91 10 10 111 3 8 121 1 10 32 1 10output
129Note
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.
题解：线段树
对于存在不同颜色的区间，每次直接暴力向下修改，直到该区间的颜色数都相同。
询问是O(logn)的应该没有问题。
但是修改的话，我并不会严格的证明，不过应该是近似O(logn)的。
#include<iostream>#include<algorithm>#include<cmath>#include<cstring>#include<cstdio>#define N 100003#define LL long longusing namespace std;int n,m,a[N];struct data{	bool pd;	LL l,num,delta;}tr[N*4];void update(int now){	tr[now].num=tr[now<<1].num+tr[now<<1|1].num;	tr[now].pd=0;	if (tr[now<<1].pd&&tr[now<<1|1].pd&&tr[now<<1].l==tr[now<<1|1].l) 	 tr[now].pd=1,tr[now].l=tr[now<<1].l; }void build(int now,int l,int r){	if (l==r) {		tr[now].l=a[l];		tr[now].num=0; tr[now].pd=1;		return;	}	int mid=(l+r)/2;	build(now<<1,l,mid);	build(now<<1|1,mid+1,r);	update(now);}LL Abs(LL x){	if (x<0) return -x;	return x;}void pushdown(int now,int l,int r){	int mid=(l+r)/2;	if (tr[now].pd) {		tr[now<<1].pd=tr[now<<1|1].pd=1;	    tr[now<<1].delta+=tr[now].delta;	    tr[now<<1|1].delta+=tr[now].delta;	    tr[now<<1].num+=tr[now].delta*(LL)(mid-l+1);	    tr[now<<1|1].num+=tr[now].delta*(LL)(r-mid);	    tr[now<<1].l=tr[now<<1|1].l=tr[now].l;	    tr[now].delta=0;	}}void qjchange(int now,int l,int r,int ll,int rr,int v){	if (ll<=l&&r<=rr) {		int mid=(l+r)/2;		if (tr[now].pd==1) {			tr[now].num+=Abs(tr[now].l-v)*(LL)(r-l+1);			tr[now].delta+=Abs(tr[now].l-v);			tr[now].l=v; 			return;		}		else qjchange(now<<1,l,mid,ll,rr,v),qjchange(now<<1|1,mid+1,r,ll,rr,v);		update(now);		return;	}	int mid=(l+r)/2;	pushdown(now,l,r);	if (ll<=mid) qjchange(now<<1,l,mid,ll,rr,v);	if (rr>mid) qjchange(now<<1|1,mid+1,r,ll,rr,v);	update(now);}LL query(int now,int l,int r,int ll,int rr){	if (ll<=l&&r<=rr) return tr[now].num;	int mid=(l+r)/2;    bool pd=false; LL t=0;    pushdown(now,l,r);	if (ll<=mid) t=query(now<<1,l,mid,ll,rr);	if (rr>mid) t+=query(now<<1|1,mid+1,r,ll,rr);	return t;}int main(){	freopen("a.in","r",stdin);	scanf("%d%d",&n,&m);	for (int i=1;i<=n;i++) a[i]=i;	build(1,1,n);	for (int i=1;i<=m;i++) {		int opt,x,y,val; data t;		scanf("%d%d%d",&opt,&x,&y);		if (opt==1) {			scanf("%d",&val);			qjchange(1,1,n,x,y,val);		}		else printf("%I64d/n",query(1,1,n,x,y));	}}