Floyd(动态规划)求解任意两点间的最短路径(图解)

Floyd算法的精髓在于动态规划的思想,即每次找最优解时都建立在上一次最优解的基础上,当算法执行完毕时一定是最优解

对于邻接矩阵w,w保存最初始情况下任意两点间的直接最短距离,但没有加入中继点进行考虑
如w[1][2]=20,即表示点1与点2的当前最短距离(直接距离)为20

对于路径矩阵path,保存了点i到点j的最短路径中下一个点的位置,
如path[1][2]=0,表示1->2的路径中的下一个点为结点0

Floyd算法对所有中继点在任意两点中进行循环遍历.即k从0-n时考虑(i->k,k->j)的路径是否小于(i->j)的路,如果小于即更新邻接矩阵w的值与path矩阵中的值,使其始终保持最短
图解如下:Floyd(动态规划)求解任意两点间的最短路径(图解)

代码用例:
Floyd(动态规划)求解任意两点间的最短路径(图解)

代码如下

点击查看代码
#include<iostream>
#include<fstream>
#include<vector>
using namespace std;
const int MAX = 999;
class Solution {
public:
	void GetPath(vector<vector<int>>vec,int n) {
		
		vector<vector<int>>path(n);

		//初始化
		for (int i = 0; i != n; i++)
			path[i].resize(n);

		for(int i=0;i!=n;i++)
			for (int j = 0; j != n; j++) {
				if (vec[i][j] != MAX)path[i][j] = j;
				else path[i][j] = -1;
			}

		for (int i = 0; i < n; i++)path[i][i] = -1;
		for(int k=0;k!=n;k++)
			for(int i=0;i!=n;i++)
				for (int j = 0; j != n; j++) {
					if (vec[i][k] + vec[k][j] < vec[i][j]) {
						vec[i][j] = vec[i][k] + vec[k][j];
						path[i][j] = path[i][k];
					}
				}
				
		for (int i = 0; i != n; i++)
		{
			cout << "\nStating from vertex: " << i << endl;
			bool flag = 0;
			for (int j = 0; j != n; j++)
				if (j != i && vec[i][j] < MAX) {
					flag = 1;
					cout << i << "->" << j << ":distance=" << vec[i][j] << ": " << i;
					int k = path[i][j];
					while (k != j) {
						cout << "->" << k;
						k = path[k][j];
					}
					cout << "->" << j << endl;
				}

			if (!flag)cout << "there's no path while starting from "<<i<<endl;
		}
	}
};
int main() {
	ifstream putIn("D:\\Input.txt", ios::in);
	int num;
	int finalCount = 0;
	putIn >> num;
	const int x = num;
	vector<vector<int>>myVector(num);
	//myVector:带权邻接矩阵
	for (int i = 0; i < num; i++)
		myVector[i].resize(num);
	for (int i = 0; i < num; i++)
		for (int j = 0; j < num; j++) {
			int temp;
			putIn >> temp;
			myVector[i][j] = temp;
		}
	Solution solution;
	cout << "Input文件中的邻接矩阵为\n";
	for (auto x : myVector) {
		for (auto y : x)cout << y << '\t';
		cout << endl;
	}
	solution.GetPath(myVector,num);
	return 0;
}

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