标签：city matrix int MST 最小 value 算法 cost cities

Minimum cost to connect all cities

• Difficulty Level : Medium

• There are n cities and there are roads in between some of the cities. Somehow all the roads are damaged simultaneously. We have to repair the roads to connect the cities again. There is a fixed cost to repair a particular road. Find out the minimum cost to connect all the cities by repairing roads. Input is in matrix(city) form, if city[i][j] = 0 then there is not any road between city i and city j, if city[i][j] = a > 0 then the cost to rebuild the path between city i and city j is a. Print out the minimum cost to connect all the cities. 

It is sure that all the cities were connected before the roads were damaged.

Examples: 

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Input : {{0, 1, 2, 3, 4},
         {1, 0, 5, 0, 7},
         {2, 5, 0, 6, 0},
         {3, 0, 6, 0, 0},
         {4, 7, 0, 0, 0}};
Output : 10

import java.io.*;
import java.util.*;

class GFG {
    public static void main (String[] args) {
        int[][] matrix = new int[][]{{0, 1, 1, 100, 0, 0},
         {1, 0, 1, 0, 0, 0},
         {1, 1, 0, 0, 0, 0},
         {100, 0, 0, 0, 2, 2},
         {0, 0, 0, 2, 0, 2},
         {0, 0, 0, 2, 2, 0}};
        System.out.println(getMinPath(matrix));
    }
    static class Edge{
        int x;
        int y;
        int value;
        Edge(int x,int y,int value){
            this.x=x;
            this.y=y;
            this.value=value;
        }
    }
    private static int getMinPath(int[][] matrix){
        int len = matrix.length;
        //1.create heap
        PriorityQueue<Edge> pq = new PriorityQueue<>((a,b)->a.value-b.value);
        //2.put the first element into heap
        Set<Integer> visited = new HashSet<>();
        visited.add(0);
        for(int i=0;i<len;i++) if(i!=0 && matrix[0][i]!=0) pq.offer(new Edge(0,i,matrix[0][i]));
        //3.while loop 
        int sum = 0;
        while(!pq.isEmpty()){
            Edge edge = pq.poll();
            if(visited.add(edge.y)){
                sum+=edge.value;
                for(int i=0;i<len;i++) if(i!=edge.y && matrix[edge.y][i]!=0) pq.offer(new Edge(edge.y,i,matrix[edge.y][i]));
            }
        }
        if(visited.size()==len) return sum;
        return -1;
    }
}

标签：city,matrix,int,MST,最小,value,算法,cost,cities
来源： https://www.cnblogs.com/cynrjy/p/15697369.html

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