Single-source shortest path problem [Dijkstra implementation]

Single-source shortest path problem [Dijkstra implementation]

I. Problems

The weighted directed graph G (E, v) is used to find the weighted minimum path from the given Source Vertex s to other vertex v.

"Shortest Path" = Minimum permission

Ii. Problem Solving:

What is the minimum path value between 1 and 5?

 

Iii. Execution Process:

 

 

If you do not quite understand the requirements for this issue, I will take you over again:

First time: from 1 --> at this time from 1 --> 3 no path all are infinite 1 --> 1 -->, then we find that the smallest group is 10, that is, 2, so we add the 2 point to the set, so the 2 point can be used as a bridge,

The second time: now we can use 1 --> 2 --> 3: 60, other 1 -->.

1 --> can be found that the minimum value should be 3 at this time, so we will add 3 to this set to do these things repeatedly, at the end, we can find that the shortest path of 1*5 is 60 (1 --> 4 --> 3 --> 5)

 

Iv. Dijkstra pseudocode:

Int Dijkstra (int s, int t) {initialized S = {empty set} d [s] = 0; other D values are positive infinity while (not t in S) {retrieve the smallest d [I]; for (all vertices not in S and adjacent to I j) if (d [J]> d [I] + cost [I] [J]) d [J] = d [I] + cost [I] [J]; ("relax" Operation) S = S + {I}; // Add the I point to the set S} return d [T];}

Why is the relaxation operation:

That is to say, if the value of 1 --> 3 is Dist [3]> Dist [2] + map [2] [3]

So Dist [3] = dits [2] + map [2] [3]

V. Code implementation:

# Include <iostream> using namespace STD; # define Max 9999999 # define Len 210int map [Len] [Len]; // The distance between two points at a certain point: int Dist [Len]; // records the shortest path length of the current point to the source point. Int mark [Len]; // set of added points // The initialized map is positive infinity void Init () {int I, j; for (I = 0; I <Len; I ++) {for (j = 0; j <Len; j ++) {map [I] [J] = max ;}}// N: How many start paths: start Point void mydijstra (int n, int start) {int I, j, Min, K; for (I = 1; I <= N; I ++) {mark [I] = 0; // no point is added to Dist [I] = map [start] [I]; // initial} mark [sta RT] = 1; // Add the start point to Dist [start] = 0; for (I = 1; I <= N; I ++) {min = max; for (j = 1; j <= N; j ++) {If (! Mark [J] & Dist [J] <min) {// retrieve the smallest Dist not in the mark [I] min = DIST [J]; k = J; // tag} If (min = max) break; Mark [k] = 1; // Add K. // perform the relaxation operation for (j = 1; j <= N; j ++) {If (! Mark [J] & Dist [J]> Dist [k] + map [k] [J]) {Dist [J] = DIST [k] + map [k] [J] ;}}} int main () {int I, j, N, line; int, b, d; CIN> N> line; // input vertex and edge Init (); for (I = 0; I <line; I ++) {CIN> A> B> D; // enter the edge weight if (Map [a] [B]> D) {map [a] [B] = map [B] [a] = D ;}} mydijstra (n, 1 ); // call the method // output the Shortest Path cout from 1 to 5 <Dist [5] <Endl; return 0 ;}

Vi. Test Data:

5 7
1 2 10
1 4 30
1 5 100
2 3 50
3 5 10
4 3 20
4 5 60

 

Result: