C. DZY Loves Colors
题目连接：
http://codeforces.com/contest/444/problem/C
Description
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
3 3
1 1 2 4
1 2 3 5
2 1 3
Sample Output
8
题意
一开始,a[i]=i,b[i]=0
然后两个操作
1.使得[l,r]的b[i]+=fabs(x-a[i]),a[i]=x
2.查询[l,r]的b[i]和
题解：
这种乱七八糟更新的，直接分块去莽一波就好了
打一个lazy标记，然后直接跑分块就好了……
当然，这个更新为定值的，其实一般都是最后可以缩成一个点什么的，用一个线段树去莽，这感觉也是可以做的，这是套路
代码
#include <bits/stdc++.h>
using namespace std;
#define lowbit(x) ((x)&(-x))
const int maxn = 1e5 + 500;
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n , m , unit , a[maxn] ;
long long extra[maxn] , lz[maxn] , ls[maxn] , b[maxn];
int main( int argc , char * argv[] ){
n = read() , m = read();
unit = sqrt( n );
for(int i = 0 ; i < n ; ++ i) a[i] = i + 1;
while( m -- ){
int op = read() , l = read() , r = read() , x;
--l;
--r;
if( op == 1 ){
x = read();
for(int i = l ; i <= r ; ){
int idx = i / unit;
int st = idx * unit;
int ed = min( n , ( idx + 1 ) * unit );
if( i == st && r >= ed - 1 ){
if( ls[idx] ){
extra[idx] += 1LL * abs( x - ls[idx] ) * ( ed - st ) ;
lz[idx] += abs( x - ls[idx] );
ls[idx] = x;
}
else{
for(int j = st ; j < ed ; ++ j){
b[j] += abs( a[j] - x );
extra[idx] += abs( a[j] - x );
a[j] = x;
}
ls[idx] = x;
}
i = ed;
}else{
if( ls[idx] ){
for(int j = st ; j < ed ; ++ j) a[j] = ls[idx];
ls[idx] = 0;
}
extra[idx] += abs( a[i] - x );
b[i] += abs( a[i] - x );
a[i] = x;
++ i;
}
}
}else{
long long res = 0;
for(int i = l ; i <= r ; ){
int idx = i / unit;
int st = idx * unit;
int ed = min( n , ( idx + 1 ) * unit );
if( i == st && r >= ed - 1 ){
res += extra[idx];
i = ed;
}else{
res += b[i] + lz[idx];
++ i;
}
}
printf("%lld\n" , res );
}
}
return 0;
}