[英]Implementing Dijkstra's algorithm using min-heap but failed
I am trying to implement Dijkstra's Algorithm
using min-heap in java but getting wrong output every time.我试图在 java 中使用最小堆实现
Dijkstra's Algorithm
,但每次都出错 output。 Here i fount the same topic in C++. Below is my graph.在这里,我在 C++ 找到了相同的主题。下面是我的图表。 Node A, which is green colored, is source and Node F, which is red colored, is destination.
绿色的节点 A 是源,红色的节点 F 是目标。 My objective is to find out the shortest path length from A to F.
我的目标是找出从 A 到 F 的最短路径长度。
Below is my code下面是我的代码
public class Dijkstra {
private static Heap heap = new Heap();
private static int[][] graph;
public Dijkstra() {
graph = new int[6][6];
/*
* The graph value assignment is just for checking the code. node A is
* referred as node 0, node B is referred as node 1 and so on. finally
* node F is referred as node 5.
*/
graph[0][0] = graph[0][1] = graph[0][3] = graph[0][4] = graph[0][5] = graph[1][0] = graph[1][1] = graph[1][4] = graph[1][5] = graph[2][2] = graph[2][5] = graph[3][0] = graph[3][3] = graph[4][0] = graph[4][1] = graph[4][4] = graph[5][0] = graph[5][1] = graph[5][2] = graph[5][5] = 0;
graph[1][2] = graph[2][1] = graph[2][3] = graph[3][2] = graph[3][4] = graph[4][3] = graph[4][5] = graph[5][4] = 1;
graph[1][3] = graph[3][1] = 3;
graph[0][2] = graph[2][0] = 4;
graph[2][4] = graph[4][2] = 5;
graph[3][5] = graph[5][3] = 8;
}
public static void main(String[] args) {
Dijkstra dij = new Dijkstra();
// Source is node A (node 0) and destination is node F (node 5)
System.out.println(dij.solve(6, 0, 5));
}
public int solve(int numOfNodes, int source, int dest) {
heap.push(source, 0);
while (!heap.isEmpty()) {
int u = heap.pop();
if (u == dest)
return heap.cost[dest];
for (int i = 0; i < numOfNodes; i++) {
if (graph[u][i] >= 0)
heap.push(i, heap.cost[u] + graph[u][i]);
}
}
return -1;
}
}
class Heap {
private int[] data;
private int[] index;
public int[] cost;
private int size;
public Heap() {
data = new int[6];
index = new int[6];
cost = new int[6];
for (int i = 0; i < 6; i++) {
index[i] = -1;
cost[i] = -1;
}
size = 0;
}
public boolean isEmpty() {
return (size == 0);
}
private void shiftUp(int i) {
int j;
while (i > 0) {
j = (i - 1) / 2;
if (cost[data[i]] < cost[data[j]]) {
// swap here
int temp = index[data[i]];
index[data[i]] = index[data[j]];
index[data[j]] = temp;
// swap here
temp = data[i];
data[i] = data[j];
data[j] = temp;
i = j;
} else
break;
}
}
private void shiftDown(int i) {
int j, k;
while (2 * i + 1 < size) {
j = 2 * i + 1;
k = j + 1;
if (k < size && cost[data[k]] < cost[data[j]]
&& cost[data[k]] < cost[data[i]]) {
// swap here
int temp = index[data[k]];
index[data[k]] = index[data[i]];
index[data[i]] = temp;
// swap here
temp = data[k];
data[k] = data[i];
data[i] = temp;
i = k;
} else if (cost[data[j]] < cost[data[i]]) {
// swap here
int temp = index[data[j]];
index[data[j]] = index[data[i]];
index[data[i]] = temp;
// swap here
temp = data[j];
data[j] = data[i];
data[i] = temp;
i = j;
} else
break;
}
}
public int pop() {
int res = data[0];
data[0] = data[size - 1];
index[data[0]] = 0;
size--;
shiftDown(0);
return res;
}
public void push(int x, int c) {
if (index[x] == -1) {
cost[x] = c;
data[size] = x;
index[x] = size;
size++;
shiftUp(index[x]);
} else {
if (c < cost[x]) {
cost[x] = c;
shiftUp(index[x]);
shiftDown(index[x]);
}
}
}
}
While running this whole code, i am getting 0 as output but one can clearly tell the cost from node A to node F is 7 (4+1+1+1 = ACDEF).在运行整个代码时,我得到 0 作为 output,但可以清楚地看出从节点 A 到节点 F 的成本是 7(4+1+1+1 = ACDEF)。 Where is the error?
错误在哪里?
You test for an existing edge using graph[u][i] >= 0
.您使用
graph[u][i] >= 0
测试现有边。 But your graph is defined to have no edge for value zero.但是您的图形被定义为没有零值边缘。 So you should change it to
所以你应该把它改成
if (graph[u][i] > 0) ...
inside method solve
.内部方法
solve
。 Another possibility is to mark non-existing edges with a value of -1
in your matrix.另一种可能性是在矩阵中用
-1
值标记不存在的边。 This would then also allow for zero-cost edges.这也将允许零成本边缘。
In the heap, you have two values: index that identifies the node, and cost that identify the distance of the node.在堆中,您有两个值:标识节点的索引和标识节点距离的成本。 You pop the cost, that is the distance, but you used it like the index to identify the node.
你弹出成本,也就是距离,但你像索引一样使用它来识别节点。
public int pop() {
int res = data[0];
...
return res;
}
and in solve():在解决()中:
int u = heap.pop();
if (u == dest)
...
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