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1. Junior Member Join Date
May 2013
Posts
6

## Building bridges

**Problem:**
There is a river that runs horizontally through an area. There are a set of cities above and below the river. Each city above the river is matched with a city below the river, and you are given this matching as a set of pairs.

You are interested in building a set of bridges across the river to connect the largest number of the matching pairs of cities, but you must do so in a way that no two bridges intersect one another.

**My Approach:**
I am sorting the first set and then finding the largest increasing subsequence from second set despite trying all test cases. I am getting wrong answer. Please help if there is something wrong with the approach or any test case i am missing.

Here is the link to the original problem: http://www.spoj.com/problems/BRIDGE/

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct dj{
int x;
int y;
}a;
inline int myf(dj dj1,dj dj2)
{
return dj1.x<dj2.x;
}
int main()
{
int n,t,L,len;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i].x);
for(int i=0;i<n;i++)
scanf("%d",&a[i].y);

sort(a,a+n,myf);

L=1;
len=1;
for(int i=1;i<n;i++)
{
L[i]=1;
for(int j=0;j<i;j++)
{
if((a[i].y>a[j].y)&&(L[i]<L[j]+1))
L[i]=L[j]+1;
}
if(len<L[i])
len=L[i];
}
printf("%d\n",len);
}
return 0;
}  Reply With Quote

2. Elite Member      Join Date
May 2009
Posts
2,413

## Re: Building bridges

Why can't you have just one bridge? That bridge would connect all cities above the river with all cities below.  Reply With Quote

3. ## Re: Building bridges Originally Posted by nuzzle Why can't you have just one bridge? That bridge would connect all cities above the river with all cities below.
The full problem description explains the scenario and the constraints.  Reply With Quote

4. Elite Member      Join Date
May 2009
Posts
2,413

## Re: Building bridges Originally Posted by 2kaud The full problem description explains the scenario and the constraints.
Well, if you get it please explain why bridging the third and fourth pairs is the correct solution?

2 5 8 10
6 4 1 2  Reply With Quote

5. Junior Member Join Date
May 2013
Posts
6

## Re: Building bridges

Originally we have 10 points.

1 2 3 4 5 6 7 8 9 10
<---- Cities on the first bank of river---->
--------------------------------------------
<--------------- River--------------->
--------------------------------------------
1 2 3 4 5 6 7 8 9 10
<------- Cities on second bank of river------->

Test Case:
2 5 8 10
6 4 1 2

We can connect 8th city on first bank to the 1st city on second bank . This gives us one bridge. explained by (8,1) pair.
secondly, we can connect 10th city on first bank to the 2nd city on second bank. This gives us 2 Non overlapping(non cutting) bridges.

But if we consider connecting 2nd city on first bank to 6th city on second bank. Then this will cut all the three remaining bridges.
Similary while connecting 5th city on first bank to 4th city on second bank.

Thefore, maximum number of overlapping bridges are 2.  Reply With Quote

6. Junior Member Join Date
May 2013
Posts
6

## Re: Building bridges

got AC Anyone having error in the above program, can check out this working code.
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct dj{
int x;
int y;
}a;
inline int myf(dj dj1,dj dj2)
{
if(dj1.x<dj2.x)
return 1;
if(dj1.x==dj2.x)
if(dj1.y<dj2.y)
return 1;
return 0;
}
int main()
{
int n,t,L,len;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i].x);
for(int i=0;i<n;i++)
scanf("%d",&a[i].y);

sort(a,a+n,myf);

L=1;
len=1;
for(int i=1;i<n;i++)
{
L[i]=1;
for(int j=0;j<i;j++)
{
if((a[i].y>=a[j].y)&&(L[i]<L[j]+1))
L[i]=L[j]+1;
}
if(len<L[i])
len=L[i];
}
printf("%d\n",len);
}
return 0;
}  Reply With Quote

7. Elite Member      Join Date
May 2009
Posts
2,413

## Re: Building bridges Originally Posted by codedhrj Thefore, maximum number of overlapping bridges are 2.
Thank you, I get it now.

I think you've come up with a nice solution strategy. First sorting the bridges according to "from" and then finding the longest increasing subsequence among the "to". That would correspond to the largest number of non-intersecting bridges.

The sorting is O(N*logN) but it looks like you have an O(N^2) algorithm for the subsequence part. According to this link, O(N*logN) is possible for the latter also,

http://en.wikipedia.org/wiki/Longest...ng_subsequence

That means a solution with a total complexity of O(N*logN) is possible for this problem.
Last edited by nuzzle; May 31st, 2013 at 02:37 AM.  Reply With Quote

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