本文主要是介绍最短路径算法，对大家解决编程问题具有一定的参考价值，需要的程序猿们随着小编来一起学习吧！

最短路径算法

Dijkstra

algorithm description

1. Select a source, and initialize the distance of other nodes to the source and finished as false
2. Find the minimum distance from dis[] and the node must be not finished
3. Update all the dis[] based on the current selected node
4. Repeat until all the nodes are finished.

time complexity: O(|V|²)

void dijkstra(int[][] graph, int source)
    {
        int[] distance = new int[V];
        Boolean[] finished = new Boolean[V];
        for (int i = 0; i < V; i++)
        {
            distance[i] = Integer.MAX_VALUE;
            finished[i] = false;
        }
        distance[source] = 0;
        //v-1个node
        for (int count = 0; count < V-1; count++)
        {
            int u = minDistance(distance, finished);
            finished[u] = true;
            for (int v = 0; v < V; v++)
                if (!finished[v] && graph[u][v]!=0 &&
                        distance[u] != Integer.MAX_VALUE &&
                        distance[u]+graph[u][v] < distance[v])
                    distance[v] = distance[u] + graph[u][v];
        }
    }
    int minDistance(int[] distance, Boolean[] finished)
    {
        int min = Integer.MAX_VALUE, min_index=-1;
        for (int v = 0; v < V; v++) {
            if (!finished[v] && distance[v] <= min) {
                min = distance[v];
                min_index = v;
            }
        }
        return min_index;
    }

Flyod

 Floyd–Warshall algorithm is an algorithm based on Dynamic Programming for finding shortest paths in a weighted graph with positive or negative edge weights (but with no negative cycles).

The Floyd–Warshall algorithm compares all possible paths through the graph between each pair of vertices.

Find shortest possible path from i to j using vertices only from the set {1, 2, ..., k} as intermediate points along the way.

time complexity: O(|V|³)

core code is pretty simple

for (int k = 0; k < n; ++k)
            for (int i = 0; i < n; ++i)
                for (int j = 0; j < n; ++j) {
                    if(A[i][j]>A[i][k]+A[j][k]){
                        A[i][j]=A[i][k]+A[j][k];
                        path[i][j]=k;
                    }
                }

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