CIELDIST - Editorial

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Contest

CAKEWALK

PREREQUISITES

Simple math, Triangle inequality

PROBLEM

We have two markets S and T. We want to build two restaurants. The markets and restaurants can be considered as points in 2D Cartesian plane. Arrange markets and the restaurants in such a way that:

• The distance between S and T is D.
• The distance between S and one of the restaurants is DS.
• The distance between T and the other restaurant is DT.
• The distance between the two restaurants is minimized.

Determine the minimum distance between the two restaurants.

QUICK EXPLANATION

To solve this kind of problem, it is best to try sketching many placements of the objects, and try spotting some easy patterns that lead to the best result. The answer to this problem is very short: max(0, D - DS - DT, DS - DT - D, DT - DS - D).

EXPLANATION

We can divide the input into 4 cases:

Case 1. D ≥ DS + DT

The optimal placement of the objects is in a straight line as follows.

```  +----------D----------+
/                       \
S        R       R        T
\      /         \      /
+-DS-+           +-DT-+
```

Here, the markets are represented by S and T while restaurants are represented by R’s. The minimum distance between the points is D - DS - DT.

Case 2. (DS + DT > D) and (D + DS > DT) and (D + DT > DS)

The lengths are said to satisfy the triangle inequality. In this case, we can always form a triangle as follows.

```  +----------D----------+
/                       \
S                         T
\                     /
\                 /
DS           DT
\         /
\     /
\ /
R
```

Here, the two restaurants are located in the exact same position and thus represented by a single R. The distance is of course 0.

Case 3. Triangle inequality does not hold and DS ≥ D + DT

The optimal placement is as follows.

```  +----------DS---------+
/                       \
S        T        R       R
\      / \      /
+-D--+   +-DT-+
```

The distance between the restaurants is DS - D - DT.

Case 4. Triangle inequality does not hold and DT ≥ D + DS

The optimal placement is as follows.

```  +----------DT---------+
/                       \
R        R       S        T
\      /\      /
+-DS-+  +-D--+
```

The distance between the restaurants is DT - D - DS.

So, the code for this solution is very simple, just considering all four cases and printing the appropriate value. However, one can actually unite the four cases in one single expression: max(0, D - DS - DT, DS - DT - D, DT - DS - D). Try to prove it yourselves!

SETTER’S SOLUTION

Can be found here.

TESTER’S SOLUTION

Can be found here.

8 Likes

Case 1. D ≥ DS + DT

The optimal placement of the objects is in a straight line as follows.

``````   +----------D----------+
/                       \
S        R       R        T
\      /         \      /
+-DS-+           +-DT-+
``````

IN this case answer would be D-(DS + DT) instead of DS - DT.

I already edited that part, hope nobody minds… It was probably a little flaw, as the answer is correct in the Quick Explanation formula…

Bruno

1 Like

Answer can also be thought in terms of 2 circles.
All cases will lie basically in 5 configurations of these circles.

28 Likes

Great job!

4 Likes

I believe the editorial says that the answer is D - DS - DT.

This is because @kuruma edited this part just after this answer

Oops… sorry then. Thanks kuruma

You’re welcome!!
We’re here to help eachother

I got exactly the same idea

in case of Ds>D+Dt
lets suppose
S=0,0 T=20,0
Ds=30 Dt=5
restaurent may be at any point within circle with radius Dt=5m with center 20,0;
for this case answer is 0 not Ds-D-T

int main()
{
int testCases;
std::cin>>testCases;
for (int i = 0; i < testCases; ++i)
{
int a,b,c;
std::cin>>a;
std::cin>>b;
std::cin>>c;
c=c-b-a;
c=c>0?c:0;
std::cout<<c<<std::endl;
}
}

//