E. DZY Loves Colorstime 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:
- 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.
Examplesinput
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.
题意
一开始,a[i]=i,b[i]=0，然后有两个操作：
1.使得[l,r]的b[i]+=fabs(x-a[i]),a[i]=x
2.查询[l,r]的b[i]和
思路一：线段树+lazy
这个应该比较轻松吧，线段树的lazy大家应该都会，这里我就不多说了。
思路二：分块+lazy
终于感觉到卿学姐之前的分块写法的漏洞了——代码量大，虽然思路清晰，但是很容易错TAT，我debug了1个小时多。
这里的关键就是如何学习lazy技巧了。
刚开始的时候我的lazy就直接保存之前输入的val，就和线段树一样那个样子更新，但是最后我写着写着发现，代码量好像有点大啊，就果断看了一下卿学姐的代码，瞬间感觉自己好傻。
这里，我们用lastval表示之前该块里面的所有的a[i]更新以后的权值，然后lazy就表示如果是重复更新的话，那么lazy[i] += abs(val – lastval[i])即可，然后这里我们再用一个sum保存整个块的val。
当询问左右区间的时候，就直接暴力即可:ans += lazy[j] + b[j]，当询问中间区间（即块）的时候，就是ans+=sum[j]
我的代码：按照卿学姐以前的习惯写的
//看看会不会爆int!数组会不会少了一维！
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker,"/STACK:102400000,102400000")
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha\n")
const int maxn = 1e5 + ;
LL a[maxn], b[maxn], add[maxn], sum[maxn], lazy[maxn], lastval[maxn];
int block, num, belong[maxn], l[maxn], r[maxn];
int n, m;void build(){
block = sqrt(n); num = n / block;
if (n % block) num++;
for (int i = ; i <= num; i++)
l[i] = (i - ) * block + , r[i] = i * block;
for (int i = ; i <= n; i++)
belong[i] = (i - ) / block + ;
}///如果lazy保存的是目前要更新的值x，那么代码的复杂度就会大大增加
///如果lazy保存的是x减pre_lazy_val，那么代码的复杂度就会大大减少，但是会多一个空间
///叫做lastval，表示保存之前的val
void update(int x, int y, int val){
if(belong[x] == belong[y]){
if (lastval[belong[x]]){
for (int i = l[belong[x]]; i <= r[belong[x]]; i++) a[i] = lastval[belong[x]];
lastval[belong[x]] = ;
}
for (int i = x; i <= y; i++){
b[i] += abs(val - a[i]); sum[belong[x]] += abs(val - a[i]);
a[i] = val;
}
return ;
}
///对x的修改
if (lastval[belong[x]]){
for (int i = l[belong[x]]; i <= r[belong[x]]; i++) a[i] = lastval[belong[x]];
lastval[belong[x]] = ;
}
for (int i = x; i <= r[belong[x]]; i++){
b[i] += abs(val - a[i]); sum[belong[x]] += abs(val - a[i]);
a[i] = val;
}
///对y的修改
if (lastval[belong[y]]){
for (int i = l[belong[y]]; i <= r[belong[y]]; i++) a[i] = lastval[belong[y]];
lastval[belong[y]] = ;
}
for (int i = l[belong[y]]; i <= y; i++){
b[i] += abs(val - a[i]); sum[belong[y]] += abs(val - a[i]);
a[i] = val;
} ///对块的修改
for (int i = belong[x] + ; i < belong[y]; i++){
if (lastval[i]){
lazy[i] += abs(val - lastval[i]);
sum[i] += 1LL * (r[i] - l[i] + ) * abs(val - lastval[i]);
lastval[i] = val;
}
else {
for (int j = l[i]; j <= r[i]; j++){
b[j] += abs(val - a[j]);
sum[i] += abs(val - a[j]);
a[j] = val;//////////////这里增加了
}
lastval[i] = val;
}
}
}LL query(int x, int y){
LL ans = ;
if (belong[x] == belong[y]){
for (int i = x; i <= y; i++) ans += b[i] + lazy[belong[x]];
return ans;
}
///对x进行操作
for (int i = x; i <= r[belong[x]]; i++)
ans += b[i] + lazy[belong[x]]; ///对y进行操作
for (int i = l[belong[y]]; i <= y; i++)
ans += b[i] + lazy[belong[y]]; ///对块进行操作
for (int i = belong[x] + ; i < belong[y]; i++)
ans += sum[i];
return ans;
}int main(){
cin >> n >> m;
for (int i = ; i <= n; i++) a[i] = i;
build();
for (int i = ; i <= m; i++){
int ty, x, y, z;
scanf("%d", &ty);
if (ty == ){
scanf("%d%d%d", &x, &y, &z);
update(x, y, z);
}
if (ty == ){
scanf("%d%d", &x, &y);
printf("%lld\n", query(x, y));
}
}
return ;
}
/*
10 10
1 1 5 3
2 1 10
1 2 7 9
2 1 10
1 10 10 11
2 1 10
1 3 8 12
2 1 10
1 1 10 3
2 1 10
*/
写完了以后我就发现自己太年轻了，果然分块还能写的更加简便啊= =。 来自：链接
#include <bits/stdc++.h>using namespace std;#define lowbit(x) ((x)&(-x))
const int maxn = 1e5 + ;inline int read(){
int x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}int n , m , unit , a[maxn] ;
long long extra[maxn] , lz[maxn] , ls[maxn] , b[maxn];int main(){
n = read() , m = read();
unit = sqrt( n );
for(int i = ; i < n ; ++ i) a[i] = i + ;
while( m -- ){
int op = read() , l = read() , r = read() , x;
--l;
--r;
if( op == ){
x = read();
for(int i = l ; i <= r ; ){
int idx = i / unit;
int st = idx * unit;//表示目前块的左端点，但是要注意，这里没有+1
int ed = min( n , (idx + ) * unit );//表示目前块的右端点
if(i == st && r >= ed - ){
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{///对于块的两端东西进行暴力
if(ls[idx]){
for(int j = st; j < ed; ++j) a[j] = ls[idx];
ls[idx] = ;
}
extra[idx] += abs(a[i] - x);
b[i] += abs(a[i] - x);
a[i] = x;
++i;
}
}
}else{
long long res = ;
for(int i = l; i <= r ;){//对于中间，全都保存在extra当中？
int idx = i / unit;
int st = idx * unit;
int ed = min(n, (idx + ) * unit);
if(i == st && r >= ed - ){
res += extra[idx];
i = ed;
}else{///对于两端进行暴力
res += b[i] + lz[idx];
++i;
}
}
printf("%lld\n" , res);
}
}
return ;
}
