What is Python's isinstance called in golang?

Starlight_ Glimmer 2020-11-17 21:08:09
python isinstance golang


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Brief introduction

Find the number of unordered triples with the same Manhattan distance between any two points .

title

Step.1

First , There is a conclusion : On the plane to the point \((x,y)\) The Manhattan distance is \(d\) The trajectory of the point , In order to \((x,y)\) Centered ,\(2d\) For the diagonal square .( It's the quadrilateral in the graph \(BCDE\)

This should look better , When the dot is sliding on a square , Ordinate \(±1\), The abscissa is also \(±1\). To be more exact , Abscissa \(A\) far \(1\), The ordinate is away from \(A\) near \(1\), Abscissa \(A\) near \(1\), The ordinate is away from \(A\) far \(1\).

But the diamond is oblique , Not easy to do , So let's spin ,“ right ”, hold \((x,y)\) become \((x+y,x-y)\).

This is probably why many of the solutions are saying that it can be transformed into Chebyshev distance .

Step.2

Another conclusion :\((x,y)\) Transform the coordinate system \(->(x+y,x-y)\), Manhattan distance in the original coordinate system = Chebyshev distance in the new coordinate system

Just a quick explanation :

At two o 'clock \((x1,y1),(x2,y2)\)

Manhattan distance in the original coordinate system \(=|x1-x2|+|y1-y2|\)

Chebyshev distance in the new coordinate system \(=max(|x1+y1-x2-y2|,|x1-y1-x2+y2|)=max(|x1-x2+y1-y2|,|x1-x2-(y1-y2)|)\)

Let's talk about it \(x1-x2\) and \(y1-y2\) The sign and absolute value of the .

Make \(x1-x2=m,y1-y2=n\)( It's easy to write

\(m,n\) Same number ,\(|m+n|=|m|+|n|,|m-n|=||m|-|n||\)

be aware , No matter what \(|m|,|n|\) How about the size relationship ,\(|m-n|=(|m|-|n|)/(|n|-|m|)\) It's all smaller than \(|m|+|n|\) Of , Obtain evidence .

\(m,n\) When there is a different sign ,\(|m+n|=||m|-|n||,|m-n|=|m|+|n|\), Empathy .

Another similar conclusion , I also add here ( This problem doesn't use ):\((x,y)->(\frac{x+y}2,\frac{x-y}2)\), Chebyshev distance of the original coordinates = Manhattan distance in new coordinates , Prove the same .


Of course , It's an algebraic way of understanding , I think it can be simpler in terms of Geometry .

It was said that , Square \(BCDE\) Is to meet the Manhattan distance for \(d\) Track of .

So if you put the trajectory right , Put every point on the square according to \((x,y)->(x+y,x-y)\) mapping ( Transform the coordinate system ), Draw one to \(2d\) For a square with a side length , And by geometric relations ( Look at the picture ) Easy to know , This square is to \(A\) Point Chebyshev distance is \(d\) The trajectory of the point .

Step.3

A conditional triple must have two dots \(x\) or \(y\) The coordinates are the same .

If you first select two points that do not meet the above requirements \(A,B\)

It's marked out in the picture \(T,U\) It's the third eligible person spot , They all satisfy the above conditions .(\(A\) The point that meets the condition is with \(A\) The side length of the center is \(3\) The square of ,\(B\) Empathy , So the point satisfying the condition is the intersection of two square trajectories )( I'm a little bit messy

Step.4

that , The qualified Sanyuan Group leader looks like this :

All the whole points on the line meet the conditions .

You can enumerate \(A,B\) spot , because \(A,B\) Of \(x/y\) The coordinates are the same , The complexity is \(O(n^3)\) Of , As for the calculation \(CD\) How many dots between , This can be prefixed and maintained .

\(x,y\) The coordinates are the same , however \(ABC,BCD,ABD,BDA\) It's going to be counted twice ( That is, the boundary points will be calculated repeatedly ), So in the second enumeration , It's OK not to include the boundary points .

Step.5

In fact, there is no need to transform the coordinate system , It's OK to take the original oblique coordinates directly , It's just a little more troublesome to write about , And it's going to rotate in four directions , But you don't have to think about the middle deduction .

But if you've seen something like this , It will be much more convenient to do questions .


►Code View

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
#define LL long long
#define N 305
#define DEL 100000
#define INF 0x3f3f3f3f
int rd()
{
int x=0,f=1;char c=getchar();
while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
while(c>='0'&&c<='9'){x=(x<<3)+(x<<1)+(c^48); c=getchar();}
return f*x;
}
int n,ans;
char mp[N][N];
int a[N<<1][N<<1],s1[N<<1][N<<1],s2[N<<1][N<<1];
int main()
{
n=rd();
for(int i=1;i<=n;i++)
{
scanf("%s",mp[i]+1);
for(int j=1;j<=n;j++)
if(mp[i][j]=='*')
a[i+j][i-j+n]=1;// To Chebyshev distance Prevent negative numbers Translation coordinates
}
n*=2;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
s1[i][j]=a[i][j]+s1[i][j-1],s2[i][j]=a[i][j]+s2[i-1][j];
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(a[i][j])// The first cow spot
for(int k=j+1;k<=n;k++)// The second cow spot And the first point i identical
if(a[i][k])
{
int d=i+(k-j);
if(d<=n) ans+=s1[d][k]-s1[d][j-1];
d=i-(k-j);
if(d>=1) ans+=s1[d][k]-s1[d][j-1];
}
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
if(a[j][i])
for(int k=j+1;k<=n;k++)
if(a[k][i])
{
int d=i+(k-j);
if(d<=n) ans+=s2[k-1][d]-s2[j][d];// We need to dig out the boundary points It's been calculated before
d=i-(k-j);
if(d>=1) ans+=s2[k-1][d]-s2[j][d];
}
printf("%d\n",ans);
return 0;
}
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本文为[Starlight_ Glimmer]所创,转载请带上原文链接,感谢

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