Toolbox: Dijkstra's

-

Toolbox: Dijkstra's Shortest Path Algorithm

What is it?

Dijkstra's is a shortest path algorithm especially optimized for sparse graphs.

Code?

Optimized (+ priority queue - needs <queue>)

struct value
{
    int dat, val; constexpr bool operator<( const value& rhs) const { return val > rhs.val; }
};

priority_queue<value> pq;
vector<value> elist[805];

void dj(int st)
{
    memset(tmp, 63, sizeof tmp);
    tmp[st] = 0;
    pq.push({st, 0});
    while(!pq.empty())
    {
        value u = pq.top(); pq.pop();
        if(tmp[u.dat] != u.val) continue;
        for(value x:elist[u.dat]) // if C++11 incomaptible, change
        {
            int alt = u.val + x.val;
            if(alt < tmp[x.dat]) { tmp[x.dat] = alt; pq.push({x.dat, alt}); }
        }
    }
}

Explanation?

Dependencies

  • <vector>, <queue>
  • C++ 11 or above
  • st, starting point
  • Edge list in elist

Outputs

  • tmp, array of distances between starting point and any point

You're welcome.

Comments

Post a Comment

Popular posts from this blog

USACO Training "subset": Subset Sums

-
USACO Training "subset": Subset Sums Problem Statement Subset Sums JRM For many sets of consecutive integers from 1 through N (1 <= N <= 39), one can partition the set into two sets whose sums are identical. For example, if N=3, one can partition the set {1, 2, 3} in one way so that the sums of both subsets are identical: {3} and {1,2} This counts as a single partitioning (i.e., reversing the order counts as the same partitioning and thus does not increase the count of partitions). If N=7, there are four ways to partition the set {1, 2, 3, ... 7} so that each partition has the same sum: {1,6,7} and {2,3,4,5} {2,5,7} and {1,3,4,6} {3,4,7} and {1,2,5,6} {1,2,4,7} and {3,5,6} Given N, your program should print the number of ways a set containing the integers from 1 through N can be partitioned into two sets whose sums are identical. Print 0 if there are no such ways. Your program must calculate the answer, not look it up from a table. PROGRAM NAME: s...
Read more

USACO 2018 Open: Talent Show

-
Image
USACO 2018 Open Talent Show In Short Given N objects with value Vi and weight Wi, a collection of elements is said to have total value-to-weight ratio as (the sum of their value) divided by (the sum of their weight). Find the best possible value-to-weight ratio of these X objects, given the restraint that the collection must have total weight at least W. Return that ratio multiplied by 1000 and floored. Analysis It would be very difficult to calculate this optimal ratio - we'd have to either try all possible collections - O(2^n), where n is at most 250, about 10 to the power of 75 calculations, OR use a greedy heuristic - get the objects with the best ratio until it satisfies the weight limit. Neither of those are too to our liking, so how about searching for it? The search space is linear and not infinite - it asks for the ratio to the closest thousandths. The search space is sorted as well - if ratio X is achievable, then ratio Y < X is also achievable, if...
Read more

UVA 787 and Big Number Multiplication

-
UVA 787 and Big Number Multiplication Problem Background: Big Numbers in this problem Big numbers are effectively storing sequence of digit chunks and directly manipulating them as if they were a large number. The most intuitive way to do this is to make an array where every element is a digit, but this is not effectively using the memory. Hence, it is logical to store numbers "millions at a time", i.e. 324876498726832746  [2]  | [1] | [0] like that. Necessities to Implement Notice that although this problem may LOOK like DP, it's really solvable by O(N^2) enumeration. Also notice that in this enumeration we will only be multiplying a large number with a normal number. Pseudocode for algorithm: CREATE LargeNumber NAME best SET -INFINITY FOR EACH i BETWEEN 0 and LENGTH OF list DO:     CREATE LargeNumber NAME product SET 1     FOR EACH j BETWEEN i AND LENGTH OF list DO         SET product TO product TIMES list[j] ...
Read more