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At first you may have a little problem with the input/output format. Taking input up to eof, breaking when getting a blank line, these things may seem new.
Here are some small pdf files, very useful to start problem solving.
The books you need to read are Knuth’s concrete math, Rosen’s discrete math, Korman’s or Shahni’s algorithm, Steven-Felix Halim’s competitive programming.
Many times you will get stuck while solving the problem, you will eat wrong answer again and again. What will you do then? The first step is to think better. If you still can’t get help. Maybe you live in a place where there are no teachers or older siblings to help you. In the age of internet it is no problem, you can take help from various forums. If you register on Jan bhai’s lightoj you will get a nice forum where many will help. Almost all online jazz including uva have forums, you will find links to them if you search on google. There you talk to big programmers, see how they practice.
In these places you will find many generous people to help you. On Fahim Bhai’s site you will find great resources, you will know what features different online jazz has. You will find a list of easy uva problems here, you can start solving them.
You understand that no matter where you are, there are all the resources on the Internet to do well in programming contests, your job is to use them and practice very well. Many times you will think “I can’t”, many people give up thinking like this, you will not give up, you are bound to do better.
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You will find many codes on the net that you can submit and get accepted but forget to do that, then it will be a huge loss, you have to be honest with yourself about it. After gaining some skill, solving good problems will be more important than solving more problems, no one will see how many problems you have solved in online judges, they will see how you do in different contests.
Finally, I would like to say that your path may not be easy, many people will discourage you, many teachers will not like you to contest, many times it will be difficult to contest along with other work, these are the reality in Bangladesh. Among these, some will give up, others will not give up on anything, only they deserve victory, you have to choose which side you will be on.
I started the contest in 2010, I have done several national and regional contests, by no means I can be called a senior contestant or a very good contestant, but in these 3 years I have gained a lot of experience, I have had the opportunity to work with many good contestants. Many times I fell behind in the ranklist and next time I recovered it and came first, still trying to improve, all in all it was a great experience to contest.
You will learn many things, beyond algorithms you will learn how to work under extreme pressure, how to recover from bad situations, how to compete with the best in the country. You may have watched the cricket game on television watching the last over with a choked breath, then after hitting a six you screamed in amazement at everyone, you will get the same experience in the contest, maybe in the last minute with a problem and beat everyone, sometimes you will raise the flag of Bangladesh in an online contest. Above all, money can’t buy this wonderful feeling.
Ending here for now. If you have any further questions, please let me know, I will be happy to help you. Good luck.
Array compression
I will discuss a very simple but useful technique today. We will do array compression using STL map. Suppose in a problem you are told that an array contains 100,000 numbers whose values range from 0 to a maximum of 1000. Now if I say a number you have to tell me which position the number is in the array. For example, suppose the input array is:
1 0 0 2 5 2 1 0 4 5 1 2
A very simple solution is to take a vector. Push i to the xth position of the vector when the number x is found in the ith position. Then after taking the input the vectors will look like:
[0]->1 2 7
[1]->0 6 10
[2]->3 5 11
[3]->empty
[4]->8
[5]->4 9
When you get which number is in which position. Now if it says where 2 is, then you loop at index 2 and print 3,5,11. A 2-D vector of size 1001 will do the job.
Now you are told that numbers can be negative and can be up to 2^30. Will this method work now? If a number is -100 or 2^30, in which index do you place its positions? Of course you can’t declare vectors that big or use negative indices. Let the input be like this:
input[]={-102,1,134565589,134565589,-102,66666668,134565589,66666668,-102,1,-2}
Note one thing, the maximum number will be 10^5 or 1 lakh. So how many different numbers can be in the array? Of course maximum 1 lakh. We will map each different number one-to-one to some smaller number. The different numbers in the above array are:
-102,1,134565589,66666668,-2
There are two ways of mapping. One is using STL map. The job of map is to assign a value to any number, string or object as we have done in the image above. In this problem we don’t need any number which is bigger or smaller so we can map it without sorting.
void compress() {
map < int, int > mymap;
int input[] = {
-102,
1,
134565589,
134565589,
-102,
66666668,
134565589,
66666668,
-102,
1,
-2
};
int assign = 0, compressed[100], c = 0, n = sizeof(input) / sizeof(int); // array size;
for (int i = 0; i < n; i++) {
int x = input[i];
if (mymap.find(x) == mymap.end()) { //x not yet compressed
mymap[x] = assign;
printf(“Mapping %d with %d\n”, x, assign);
assign++;
}
x = mymap[x];
compressed[c++] = x;
}
printf(“Compressed array: “);
for (int i = 0; i < n; i++) printf(“%d “, compressed[i]);
puts(“”);
}
The output of the code will be:
Mapping -102 with 0
Mapping 1 with 1
Mapping 134565589 with 2
Mapping 66666668 with 3
Mapping -2 with 4
Compressed array: 0 1 2 2 0 3 2 3 0 1 4
Whenever we get a new number in the input array we assign it a value and replace the original value with the map value in another array. Then we can use the new array to solve the problem. If the query says find where -2 is then you will actually find where 4 is because mymap[-2]=4.
Many times it is necessary to know which number is greater than who after compression. Like it can tell you that after taking a value, you can’t take any smaller values. How to store this information In the above method the value assigned to -2 is larger than 1. If you don’t want that, sort the different values into another array first.
sorted[]={-102,-2,1,66666668,134565589}
This time do the mapping using the sorted array. In this case the actual number at that position in the sorted array will be its mapped value.
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If you understand this then surely you can compress the array by doing binary search without map. Loop over the input array, for each x, see where x is in the sorted array. You put that value in another array:
input[]={-102,1,134565589,134565589,-102,66666668,134565589,66666668,-102,1,-2}
compressed_input[]={0,2,4,4,0,3,4,3,0,2,1}
At the time of query, you have to work out the value to compress by binary search.
You can compress the map or binary search however you want. Each time map is accessed logn complexity is equal to binary search. Different classes in stl, map is a bit slow for using dynamic memory, but if the time limit is not too tight, it won’t be a problem. Using map makes coding easier.
Some graph problems input nodes and edges as strings. For example, suppose the 3 edges of the graph are:
3
BAN AUS
AUS SRI
SRI BAN
String can also be mapped to integer in map. Assign a value to each node using this facility:
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