FloydWarshall 无限

Question

我是编程业余爱好者。我正在尝试展示该算法的每个步骤，但我想在每个 D 矩阵上将 1000 作为 INF 呈现。

需要更多行才能提交需要更多行提交需要更多行提交需要更多行提交需要更多行提交需要更多行提交需要更多行提交需要更多行提交需要更多行提交需要更多行提交需要更多行提交

#include <iostream>
#include <limits>
#define INF 1000

using namespace std;

/* my example matrix is 
0  3  inf  7
8  0   2  inf
5 inf  0   1
2 inf inf  0

i want to write that each D matrix show 1000 as INF
*/

void FloydWarshall(int **dist, int V){

int i,j,k;

for(k = 0; k < V; k++){
cout<<"D"<<k<<" matrix is: "<<endl;

for(i = 0; i < V; i++){

    for(j = 0; j < V; j++){
        cout<<dist[i][j]<<"     ";
        if(dist[i][k] != INF && dist[j][k] != INF && dist[i][j] > (dist[i][k] + dist[k][j]))
                dist[i][j] = dist[i][k]+dist[k][j];  
                          } 
                cout<<"  "<<endl;
                      }
                cout<<"  "<<endl;
                  }     


for(i = 0; i < V; i++){ 
    for(j = 0; j < V; j++){
        cout<<" "<<endl;
        cout<<"Shortest path between "<<i<<" and "<<j<<" is : "<<endl;
        if(dist[i][j]==INF)
                cout<<"INF"<<endl;
        else
                cout<<dist[i][j]<<endl;
                          } 
                      }    

   }

int main(){


int i,j,n;
int **dist;
int *cost;

cout<<"Please, enter the number of vertices: "<<endl;
cin>>n;

dist = new int*[n];

for(i = 0;i < n; i++){
dist[i] = new int[n];
}

cout<<"Please, enter the adjacency matrix: "<<endl;
cout<<"Do not forget "<<INF<<" if there is no connection between two vertices"<<endl;
for(i = 0; i < n; i++){
    for(j = 0; j < n; j++){
        cin>>dist[i][j];
        if (dist[i][j] == 0 && i != j){
            dist[i][j] = INF;
 }
 }
 }
cout<<"  "<<endl;

FloydWarshall(dist,n);
cout<<" "<<endl;
cout<<"     The Distance Matrix is:     "<<endl;
  for(i=0;i<n;i++)
 {
  for(j=0;j<n;j++)
 {
      cout<<dist[i][j]<<"     ";
 }
cout<<"\n";
 }
return 0;
 }

Answer 1

if (variable >= 1000) { printf("INF"); }

我选择了一个随机变量名，但这个逻辑应该符合你的要求。

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