Floyd-Warshall STP實施

Question

我遇到了Floyd-Warshall算法的問題。 如果輸入具有多於4個頂點，則它不起作用。 為了制作第二維動態數組，我制作動態數組[N * N]並訪問A [i，j] = A [（i-1）* N + j]

void floyd_Algorithm(fstream &F2,int N,int matrixGraph[],int matrixP[])
{
for (int k=1; k<=N; k++)
    for (int i=1; i<=N; i++)
        for (int j=1; j<=N; j++)
        {
            if (matrixGraph[(i-1)*N+j] > matrixGraph[(i-1)*N+k] + matrixGraph[(k-1)*N+j])
            {
                matrixGraph[(i-1)*N+j] =  matrixGraph[(i-1)*N+k] + matrixGraph[(k-1)*N+j];
                matrixP[(i-1)*N+j] = k ;
            }
}

這里是4個頂點矩陣的輸入

4

0 10 6 2 | 10 0 5 3 | 6 5 0 1 | 2 3 1 0 |

產量

0 5 3 2
5 0 4 3
3 4 0 1
2 3 1 0

1 4 4 1
4 2 4 2
4 4 3 3
4 4 4 4

7個頂點矩陣輸入

7

0 3 6 0 0 0 0 | 3 0 2 4 0 0 0 | 6 2 0 1 4 2 0 | 0 4 1 0 2 0 4 | 0 0 4 2 0 2 1 | 0 0 2 0 2 0 1 | 0 0 0 4 1 1 0 |

產量

0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0

1 5 7 1 1 1 1
5 2 7 5 2 2 2
7 7 3 7 7 7 3
4 5 7 4 1 4 1
5 5 7 1 5 1 1
6 6 7 6 1 6 1
7 7 7 1 1 1 7

Answer 1

你的指數有問題。

如果你的頂點是1 <= v <= N的arngle，那么i和j之間的路徑應該是矩陣[（i-1）* N + j-1]

為避免錯誤，您可能應將頂點保持在0 <= v <N的范圍內，因為（int i = 0; i <N; i ++），矩陣[i * N + j]

Answer 2

#include <iostream>
#include <iomanip>
#include <stdlib.h>
#include <fstream>
#include <vector>
using namespace std;

void floyd_Algorithm(fstream &F2,int N,int matrixGraph[],int matrixP[])
{
for (int k=0; k<N; k++)
    for (int i=0; i<N; i++)
        for (int j=0; j<N; j++)
        {
            if (matrixGraph[i*N+j] > matrixGraph[i*N+k] + matrixGraph[k*N+j])
            {
                matrixGraph[i*N+j] =  matrixGraph[i*N+k] + matrixGraph[k*N+j];
                matrixP[i*N+j] = k ;
            }
}
cout << "Ma tran duong di ngan nhat sau khi xu ly :\n";
for (int i=0; i<N; i++)
{
    cout <<"\n";
    for (int j=0; j<N; j++)
        cout << matrixGraph[i*N+j] <<"   ";
}
cout << "\nMa tran luu dinh sau khi xu ly :\n";
for (int i=0; i<N; i++)
{
    cout <<"\n";
    for (int j=0; j<N; j++)
        cout << matrixP[i*N+j] <<"   ";
}
}
void dijkstra_Algorithm(fstream &F2,int N,int matrixGraph[],int matrixP[])
{

}
int main()
{
int n,t,c;
fstream f1,f2;
f1.open("D:\\Input3.INP",ios::in);
f2.open("D:\\Output3.OUT",ios::out);
f1 >> n;
int matrix_graph[n*n];
for (int i=0; i<n; i++)
    for (int j=0; j<n; j++)
    {
        f1 >> matrix_graph[i*n+j];
    }
int matrix_p[n*n];
for (int i=0; i<n; i++)
    for (int j=0; j<n; j++)
    {
        matrix_p[i*n+j] = i;
    }
cout << "Hay nhap thuat toan muon su dung :\n";
cout << "1 : Floyd-Warshall Algorithm\n";
cout << "2 : Dijkstra Algorithm\n";
while ((t!=1)&&(t!=2))
{
    cout << "Enter : ";
    cin >> t;
}
cout << "Ma tran trong so da nhap la :\n";
for (int i=0; i<n; i++)
{
    cout <<"\n";
    for (int j=0; j<n; j++)
        cout << matrix_graph[i*n+j] <<"   ";
}
cout << "\n";
switch (t)
{
case 1 :
    floyd_Algorithm(f2,n,matrix_graph,matrix_p);
    break;
case 2 :
    dijkstra_Algorithm(f2,n,matrix_graph,matrix_p);
    break;
}
return 0;
}

我的完整代碼在這里，改變索引仍然會產生相同的結果