C. 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:
- 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.
Examples
input
Copy
3 31 1 2 41 2 3 52 1 3
output
Copy
8
input
Copy
3 41 1 3 42 1 12 2 22 3 3
output
Copy
321
input
Copy
10 61 1 5 31 2 7 91 10 10 111 3 8 121 1 10 32 1 10
output
Copy
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.
题目意思：给出一个序列初始值为1~n，我们能对它有两种操作，1.l~r区间颜色染成x 然后我们能得到 abs(x-col[l])的色差 2.求l~r色差和
思路：线段树的区间更新，我们维护一个颜色，如果颜色相同时才能成段更新
#include <bits/stdc++.h>
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxn = 100005;
ll col[maxn<<2],lazy[maxn<<2],sum[maxn<<2];
void push_up(int rt)
{
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
col[rt]=(col[rt<<1]==col[rt<<1|1]?col[rt<<1]:0);
}
void build(int l,int r,int rt)
{
if(l==r)
{
col[rt]=l;
return ;
}
int m=(l+r)>>1;
build(lson);
build(rson);
push_up(rt);
}
void push_down(int rt,int len)
{
if(col[rt])
{
col[rt<<1]=col[rt<<1|1]=col[rt];
sum[rt<<1]+=lazy[rt]*(ll)(len-(len>>1));
sum[rt<<1|1]+=lazy[rt]*(ll)(len>>1);
lazy[rt<<1]+=lazy[rt];
lazy[rt<<1|1]+=lazy[rt];
col[rt]=lazy[rt]=0;
}
}
void updata(int l,int r,int rt,int L,int R,ll val)
{
if(L<=l&&r<=R&&col[rt])
{
sum[rt]+=abs(col[rt]-val)*(ll)(r-l+1);
lazy[rt]+=abs(col[rt]-val);
col[rt]=val;
return ;
}
push_down(rt,r-l+1);
int m=(l+r)>>1;
if(L<=m) updata(lson,L,R,val);
if(R>m) updata(rson,L,R,val);
push_up(rt);
}
ll query(int l,int r,int rt,int L,int R)
{
if(L<=l&&r<=R)
{
return sum[rt];
}
push_down(rt,r-l+1);
ll ans=0;
int m=(l+r)>>1;
if(L<=m) ans+=query(lson,L,R);
if(R>m) ans+=query(rson,L,R);
return ans;
}
int main()
{
int n,m,fg,l,r;
ll x;
scanf("%d %d",&n,&m);
build(1,n,1);
while(m--)
{
scanf("%d",&fg);
if(fg==1)
{
scanf("%d %d %lld",&l,&r,&x);
updata(1,n,1,l,r,x);
}
else
{
scanf("%d %d",&l,&r);
printf("%lld\n",query(1,n,1,l,r));
}
}
}
PS:摸鱼怪的博客分享，欢迎感谢各路大牛的指点~
原文地址：https://www.cnblogs.com/MengX/p/9451934.html