标签：sample seconds get begin lap 线段树 examples 分享 tin
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:
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~
codeforces - 444c DZY Loves Colors(线段树+染色)
标签：sample seconds get begin lap 线段树 examples 分享 tin
原文地址：https://www.cnblogs.com/MengX/p/9451934.html
