2 seconds
256 megabytes
standard input
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:
- 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?
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 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
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.
题意：区间修改颜色，colorfulness+=颜色差的绝对值，区间查询colorfulness
线段树区间更新和区间查询
col记录颜色，0说明颜色不同；sum是colorfulness；lazy记录增量(colorfulness的增量)
更新的时候遇到相同颜色的被包含的区间直接更新，否则下传标记更新孩子然后合并
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define m ((l+r)>>1)
#define lson o<<1,l,m
#define rson o<<1|1,m+1,r
#define lc o<<1
#define rc o<<1|1
using namespace std;
typedef long long ll;
const int N=5e5+,INF=2e9+;
inline int read(){
char c=getchar();int x=,f=;
while(c<''||c>''){if(c=='-')f=-;c=getchar();}
while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
return x*f;
}
int n,q,op,ql,qr,x;
struct node{
ll sum,lazy,col;
}t[N<<];
void merge(int o){
t[o].col=t[lc].col!=t[rc].col?:t[lc].col;
t[o].sum=t[lc].sum+t[rc].sum;
}
void build(int o,int l,int r){
if(l==r) t[o].col=l;
else{
build(lson);
build(rson);
}
}
void pushDown(int o,int len){
if(t[o].col){
t[lc].col=t[rc].col=t[o].col;
t[lc].lazy+=t[o].lazy;
t[rc].lazy+=t[o].lazy;
t[lc].sum+=t[o].lazy*(len-(len>>));
t[rc].sum+=t[o].lazy*(len>>); t[o].col=t[o].lazy=;
}
}
void update(int o,int l,int r,int ql,int qr,ll v){//printf("update %d %d %d\n",o,l,r);
if(ql<=l&&r<=qr&&t[o].col){
t[o].sum+=(r-l+)*abs(v-t[o].col);
t[o].lazy+=abs(v-t[o].col);
t[o].col=v;
}else{
pushDown(o,r-l+);
if(ql<=m) update(lson,ql,qr,v);
if(m<qr) update(rson,ql,qr,v);
merge(o);
}
}
ll query(int o,int l,int r,int ql,int qr){
if(ql<=l&&r<=qr) return t[o].sum;
else{
pushDown(o,r-l+);
ll ans=;
if(ql<=m) ans+=query(lson,ql,qr);
if(m<qr) ans+=query(rson,ql,qr);
return ans;
}
}
int main(){
n=read();q=read();
build(,,n);
for(int i=;i<=q;i++){
op=read();ql=read();qr=read();
if(op==){x=read();update(,,n,ql,qr,x);}
else printf("%I64d \n",query(,,n,ql,qr));
}
}