给定一个有向图，其中可能包含循环，每个边都有权重，因此任务是找到从给定源顶点s到目标顶点t的任何简单路径的最小成本。简单路径是从一个顶点到另一个顶点的路径，这样就不会多次访问一个顶点。如果没有简单的路径，则返回INF(infinite)。
该图以邻接矩阵表示形式给出，其中graph [i] [j]的值指示从顶点i到顶点j的边的权重，值INF(infinite)指示从i到j的边。
例子：
Input : V = 5, E = 6
s = 0, t = 2
graph[][] = 0 1 2 3 4
0 INF -1 INF 1 INF
1 INF INF -2 INF INF
2 -3 INF INF INF INF
3 INF INF -1 INF INF
4 INF INF INF 2 INF
Output : -3
Explanation :
The minimum cost simple path between 0 and 2 is given by:
0 -----> 1 ------> 2 whose cost is (-1) + (-2) = (-3).
Input : V = 5, E = 6
s = 0, t = 4
graph[][] = 0 1 2 3 4
0 INF -7 INF -2 INF
1 INF INF -11 INF INF
2 INF INF INF INF INF
3 INF INF INF 3 -4
4 INF INF INF INF INF
Output : -6
Explanation :
The minimum cost simple path between 0 and 2 is given by:
0 -----> 3 ------> 4 whose cost is (-2) + (-4) = (-6). 方法 ：
解决上述问题的主要思想是使用深度优先搜索的修改版本遍历从s到t的所有简单路径，并在其中找到最小成本路径。关于DFS的一项重要发现是它一次遍历一条路径，因此我们可以通过在离开节点之前将节点标记为未访问来使用DFS独立遍历单独的路径。
一个简单的解决方案是从s开始，转到所有相邻的顶点，然后对其他相邻的顶点进行递归操作，直到到达目的地。即使图形中存在负权重循环或自身边缘，该算法也将起作用。
下面是上述方法的实现：
C++
// C++ code for printing Minimum Cost
// Simple Path between two given nodes
// in a directed and weighted graph
#include
using namespace std;
// Define number of vertices in
// the graph and infinite value
#define V 5
#define INF INT_MAX
// Function to do DFS through the nodes
int minimumCostSimplePath(int u, int destination,
bool visited[], int graph[][V])
{
// check if we find the destination
// then further cost will be 0
if (u == destination)
return 0;
// marking the current node as visited
visited[u] = 1;
int ans = INF;
// traverse through all
// the adjacent nodes
for (int i = 0; i < V; i++) {
if (graph[u][i] != INF && !visited[i]) {
// cost of the further path
int curr = minimumCostSimplePath(i,
destination, visited, graph);
// check if we have reached the destination
if (curr < INF) {
// Taking the minimum cost path
ans = min(ans, graph[u][i] + curr);
}
}
}
// unmarking the current node
// to make it available for other
// simple paths
visited[u] = 0;
// returning the minimum cost
return ans;
}
// driver code
int main()
{
// initialising the graph
int graph[V][V];
for (int i = 0; i < V; i++) {
for (int j = 0; j < V; j++) {
graph[i][j] = INF;
}
}
// marking all nodes as unvisited
bool visited[V] = { 0 };
// initialising the edges;
graph[0][1] = -1;
graph[0][3] = 1;
graph[1][2] = -2;
graph[2][0] = -3;
graph[3][2] = -1;
graph[4][3] = 2;
// source and destination
int s = 0, t = 2;
// marking the source as visited
visited[s] = 1;
cout << minimumCostSimplePath(s, t,
visited, graph);
return 0;
} Java
// Java code for printing Minimum Cost
// Simple Path between two given nodes
// in a directed and weighted graph
import java.util.*;
import java.lang.*;
class GFG{
// Define number of vertices in
// the graph and infinite value
static int V = 5;
static int INF = Integer.MAX_VALUE;
// Function to do DFS through the nodes
static int minimumCostSimplePath(int u, int destination,
boolean visited[],
int graph[][])
{
// Check if we find the destination
// then further cost will be 0
if (u == destination)
return 0;
// Marking the current node as visited
visited[u] = true;
int ans = INF;
// Traverse through all
// the adjacent nodes
for(int i = 0; i < V; i++)
{
if (graph[u][i] != INF && !visited[i])
{
// Cost of the further path
int curr = minimumCostSimplePath(i,
destination, visited, graph);
// Check if we have reached the
// destination
if (curr < INF)
{
// Taking the minimum cost path
ans = Math.min(ans, graph[u][i] + curr);
}
}
}
// Unmarking the current node
// to make it available for other
// simple paths
visited[u] = false;
// Returning the minimum cost
return ans;
}
// Driver code
public static void main(String[] args)
{
// Initialising the graph
int graph[][] = new int[V][V];
for(int i = 0; i < V; i++)
{
for(int j = 0; j < V; j++)
{
graph[i][j] = INF;
}
}
// Marking all nodes as unvisited
boolean visited[] = new boolean[V];
// Initialising the edges;
graph[0][1] = -1;
graph[0][3] = 1;
graph[1][2] = -2;
graph[2][0] = -3;
graph[3][2] = -1;
graph[4][3] = 2;
// Source and destination
int s = 0, t = 2;
// Marking the source as visited
visited[s] = true;
System.out.println(minimumCostSimplePath(s, t,
visited, graph));
}
}
// This code is contributed by offbeatPython3
# Python3 code for printing Minimum Cost
# Simple Path between two given nodes
# in a directed and weighted graph
import sys
V = 5
INF = sys.maxsize
# Function to do DFS through the nodes
def minimumCostSimplePath(u, destination,
visited, graph):
# Check if we find the destination
# then further cost will be 0
if (u == destination):
return 0
# Marking the current node as visited
visited[u] = 1
ans = INF
# Traverse through all
# the adjacent nodes
for i in range(V):
if (graph[u][i] != INF and not visited[i]):
# Cost of the further path
curr = minimumCostSimplePath(i, destination,
visited, graph)
# Check if we have reached the destination
if (curr < INF):
# Taking the minimum cost path
ans = min(ans, graph[u][i] + curr)
# Unmarking the current node
# to make it available for other
# simple paths
visited[u] = 0
# Returning the minimum cost
return ans
# Driver code
if __name__=="__main__":
# Initialising the graph
graph = [[INF for j in range(V)]
for i in range(V)]
# Marking all nodes as unvisited
visited = [0 for i in range(V)]
# Initialising the edges
graph[0][1] = -1
graph[0][3] = 1
graph[1][2] = -2
graph[2][0] = -3
graph[3][2] = -1
graph[4][3] = 2
# Source and destination
s = 0
t = 2
# Marking the source as visited
visited[s] = 1
print(minimumCostSimplePath(s, t, visited, graph))
# This code is contributed by rutvik_56C#
// C# code for printing Minimum Cost
// Simple Path between two given nodes
// in a directed and weighted graph
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{
// Define number of vertices in
// the graph and infinite value
static int V = 5;
static int INF = int.MaxValue;
// Function to do DFS through the nodes
static int minimumCostSimplePath(int u, int destination,
bool[] visited, int[, ] graph)
{
// Check if we find the destination
// then further cost will be 0
if (u == destination)
return 0;
// Marking the current node as visited
visited[u] = true;
int ans = INF;
// Traverse through all
// the adjacent nodes
for (int i = 0; i < V; i++)
{
if (graph[u, i] != INF && !visited[i])
{
// Cost of the further path
int curr = minimumCostSimplePath(i, destination,
visited, graph);
// Check if we have reached the
// destination
if (curr < INF)
{
// Taking the minimum cost path
ans = Math.Min(ans, graph[u, i] + curr);
}
}
}
// Unmarking the current node
// to make it available for other
// simple paths
visited[u] = false;
// Returning the minimum cost
return ans;
}
// Driver code
public static void Main(string[] args)
{
// Initialising the graph
int[, ] graph = new int[V, V];
for (int i = 0; i < V; i++)
{
for (int j = 0; j < V; j++)
{
graph[i, j] = INF;
}
}
// Marking all nodes as unvisited
bool[] visited = new bool[V];
// Initialising the edges;
graph[0, 1] = -1;
graph[0, 3] = 1;
graph[1, 2] = -2;
graph[2, 0] = -3;
graph[3, 2] = -1;
graph[4, 3] = 2;
// Source and destination
int s = 0, t = 2;
// Marking the source as visited
visited[s] = true;
Console.WriteLine(minimumCostSimplePath(s, t, visited, graph));
}
}
// This code is contributed by sanjeev2552输出：
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