Shortest path algorithm learning Summary

Dijkstra shortest path algorithm:

The advantage of Dijkstra algorithm is that it can find the shortest distance from one point to all other points;

Input:

5 7
1 2 10
1 3 20
1 5 30
2 5 10
2 3 5
4 5 20
4 3 30

Output:

0 10 15 40 20 // This is the shortest distance from 1 to all vertices in this tree.

1 # include <iostream> 2 # include <cstring> 3 using namespace STD; 4 const int n = 1100; 5 const int INF = 1000000; 6 int map [N] [N]; 7 // int P [N]; 8 int vis [N]; 9 int low [N]; 10 int n, m; // represents the number of points and the number of routes 11 void Dijkstra (INT s) 12 {13 for (INT I = 1; I <= N; I ++) 14 {15 low [I] = map [s] [I]; 16} 17 low [s] = 0; 18 vis [s] =-1; 19 int V; 20 For (INT I = 1; I <n; I ++) 21 {22 int min = inf; 23 for (Int J = 1; j <= N; j ++) 24 {25 if (vis [J]! =-1 & low [J] <min) 26 {27 min = low [J]; 28 v = J; 29} 30} 31 vis [v] =-1; 32 For (Int J = 1; j <= N; j ++) 33 {34 if (vis [J]! =-1 & low [J]> low [v] + map [J] [v]) 35 low [J] = low [v] + map [J] [v]; // continuously update the shortest distance between two points 36} 37} 38} 39 int main () 40 {41 while (CIN> N> m) 42 {43 // memset (MAP, 0, sizeof (MAP); 44 for (INT I = 1; I <= N; I ++) 45 for (Int J = 1; j <= N; j ++) 46 map [I] [J] = inf; 47 memset (VIS, 0, sizeof (VIS); 48 int A, B, C; 49 for (int I (1); I <= m; I ++) 50 {51 scanf ("% d", & A, & B, & C ); 52 map [a] [B] = map [B] [a] = C; 53} 54 Dijkstra (1); 55 for (INT I = 1; I <= N; I ++) 56 {57 cout <low [I] <""; 58} 59 cout <Endl; 60} 61 Return 0; 62}

Floyd algorithm:

The Floyd algorithm is not suitable for processing large amounts of data because it is relatively free of space and time;

1 # include <iostream> 2 using namespace STD; 3 const int n = 300; // this field should not be too large, otherwise 4 const int INF = 11111111 will be reported; 5 Int main () 6 {7 int dis [N] [N]; 8 int n, m; // The number of points and the number of routes respectively 9 While (CIN> N> m) 10 {11 for (INT I = 1; I <= N; I ++) 12 for (Int J = 1; j <= N; j ++) 13 dis [I] [J] = inf; 14 for (INT I = 1; I <= N; I ++) 15 dis [I] [I] = 0; 16 int A, B, C; 17 for (INT I = 1; I <= m; I ++) 18 {19 scanf ("% d", & A, & B, & C ); 20 if (DIS [a] [B]> C) 21 dis [a] [B] = di S [B] [a] = C; 22} 23 for (int K = 1; k <= N; k ++) 24 for (INT I = 1; I <= N; I ++) 25 {26 for (Int J = I + 1; j <= N; j ++) 27 {28 If (DIS [I] [k]! = Inf & dis [k] [J]! = Inf & dis [I] [J]> dis [I] [k] + dis [k] [J]) 29 dis [J] [I] = dis [I] [J] = dis [I] [k] + dis [k] [J]; 30} 31} 32 int W, U; 33 scanf ("% d", & W, & U ); 34 if (DIS [u] [W] = inf) printf ("no solution \ n"); 35 else printf ("% d \ n ", dis [u] [W]); 36} 37 return 0; 38}