นี่คือโปรแกรม C++ หา SSSP (Single Source Shortest Path) ใน DAG (Directed Acyclic Graphs) ใช้ Dijkstra Algorithm หาเส้นทางสั้นที่สุดจากโหนดแรกไปทุกโหนดอื่น พร้อมแสดงความยาวเส้นทาง
อัลกอริทึม
Begin
รับองค์ประกอบของกราฟ
function shortestpath():
เริ่มต้นตัวแปร
a[i] = 1
d[i] = 0
s[i].from = 0
ลูป for i = 0 ถึง 3
if b[0][i] == 0 continue
else d[i] = b[0][i], s[i].from = 0
ลูป while (c < 4)
min = INFINITY
for i = 0 ถึง 3
if min <= d[i] หรือ d[i] == 0 หรือ a[i] == 1 continue
else if min > d[i] min = d[i]
for k = 0 ถึง 3
if (min == d[k]) t = k, break
else continue
a[t] = 1
for j = 0 ถึง 3
if a[j] == 1 หรือ b[t][j] == 0 continue
else if a[j] != 1
if d[j] > (d[t] + b[t][j])
d[j] = d[t] + b[t][j]
s[i].from = t
เพิ่ม c
ลูป for i = 0 ถึง 3 พิมพ์ minimum cost
End
ตัวอย่างโค้ด
#include <iostream>
using namespace std;
#define INFINITY 9999
struct node {
int from;
} s[4];
int c = 0;
void djikstras(int *a, int b[][4], int *d) {
int i = 0, j, min, t;
a[i] = 1;
d[i] = 0;
s[i].from = 0;
for (i = 0; i < 4; i++) {
if (b[0][i] == 0) continue;
else {
d[i] = b[0][i];
s[i].from = 0;
}
}
while (c < 4) {
min = INFINITY;
for (i = 0; i < 4; i++) {
if (min <= d[i] || d[i] == 0 || a[i] == 1) continue;
else if (min > d[i]) min = d[i];
}
for (int k = 0; k < 4; k++) {
if (min == d[k]) {
t = k;
break;
}
else continue;
}
a[t] = 1;
for (j = 0; j < 4; j++) {
if (a[j] == 1 || b[t][j] == 0) continue;
else if (a[j] != 1) {
if (d[j] > (d[t] + b[t][j])) {
d[j] = d[t] + b[t][j];
s[i].from = t;
}
}
}
c++;
}
for (int i = 0; i < 4; i++) {
cout << "from node " << s[i].from << " cost is:" << d[i] << endl;
}
}
int main() {
int a[4];
int d[4];
for (int k = 0; k < 4; k++) d[k] = INFINITY;
for (int i = 0; i < 4; i++) a[i] = 0;
int b[4][4];
for (int i = 0; i < 4; i++) {
cout << "enter values for " << (i + 1) << " row" << endl;
for (int j = 0; j < 4; j++) cin >> b[i][j];
}
djikstras(a, b, d);
}
ผลลัพธ์
enter values for 1 row 0 1 3 2 enter values for 2 row 2 1 3 0 enter values for 3 row 2 3 0 1 enter values for 4 row 1 3 2 0 from node 0 cost is:0 from node 0 cost is:1 from node 0 cost is:3 from node 0 cost is:2