[英]Dijkstra's Algorithm using Adjacency Matrix not finding correct distance/path from each node to every other node
我正在嘗試編寫一個程序,該程序使用鄰接矩陣構建圖,然后使用Dijkstra的算法找到從每個節點到每個其他節點的最短路徑。 我的程序當前無法每次都能找到正確的最短路徑。 我還需要跟蹤路徑,但是我不確定從哪里開始。
class GraphD
{
public:
GraphD();
void buildGraph(ifstream &infile);
void insertEdge(int from, int to, int distance);
void findShortestPath();
private:
static const int MAXNODES = 101;
static const int infinity = 2147483647;
struct TableType
{
bool visited;
int dist;
int path;
};
int C[MAXNODES][MAXNODES]; // holds adjacency matrix
int size;
TableType T[MAXNODES][MAXNODES]; // for dijkstra's algorithm
};
#include "GraphD.h"
GraphD::GraphD()
{
size = 0;
for(int i = 1; i < MAXNODES; i++)
{
for(int j = 1; j < MAXNODES; j++)
{
C[i][j] = infinity;
T[i][j].dist = infinity;
T[i][j].visited = false;
T[i][j].path = 0;
}
}
}
void GraphD::buildGraph(ifstream &infile)
{
string line;
if(getline(infile, line))
{
size = atoi(line.c_str());
for(int i = 1; i <= size; i++)
{
getline(infile, line);
data[i] = line;
}
int vertex1, vertex2, distance;
while(getline(infile, line))
{
stringstream edge(line);
edge >> vertex1 >> vertex2 >> distance;
if(vertex1 == 0)
break;
insertEdge(vertex1, vertex2, distance);
}
for(int i = 1; i <= size; i++)
{
C[i][i] = 0;
}
}
}
void GraphD::insertEdge(int from, int to, int distance)
{
C[from][to] = distance;
}
void GraphM::findShortestPath()
{
for(int source = 1; source <= size; source++)
{
T[source][source].dist = 0;
for(int i = 1; i <= size; i++)
{
int v = 0;
int shortestDistance = infinity;
for(int j = 1; j <= size; j++)
{
if((C[source][j] < shortestDistance) && !T[source][j].visited)
{
shortestDistance = C[source][j];
v = j;
}
}
T[source][v].visited = true;
for(int w = 1; w <= size; w++)
{
if(!T[v][w].visited)
{
T[v][w].dist = min(T[v][w].dist, T[source][v].dist + C[v][w]);
}
}
}
}
}
將無窮大值設置為等於1'000'000'000(或類似的值),因為當您使用MAX_INT值時,此處會出現整數溢出
T[v][w].dist = min(T[v][w].dist, T[source][v].dist + C[v][w]);
另外,我認為您應該替換以下代碼部分
if(!T[v][w].visited)
{
T[v][w].dist = min(T[v][w].dist, T[source][v].dist + C[v][w]);
}
到下一個
if(!T[source][w].visited)
{
T[source][w].dist = min(T[source][w].dist, T[source][v].dist + C[v][w]);
}
因為您需要找到從頂點源到頂點W的距離,而不是V。
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