Find the minimum weighted cycle in an undirected graph can be found from. An example from  is shown here to well explain this problem.
Q&A from 
“Let G be a directed weighted graph with no negative cycles. Design an algorithm to find a minimum weight cycle in G that runs with a time complexity of O(|V|³).”
The above is a question I have been working on as part of my coursework. When I first read it, I immediately thought that the Floyd-Warshall algorithm would solve this problem — mainly because F-W runs in O(|V|³) time and it works for both positive and negative weighted graphs with no negative cycles. However, I soon remembered that F-W is designed to find the shortest path of a graph, not a minimum weight cycle.
Am I on the right track with this question? Would it be possible to modify the Floyd-Warshall algorithm to find a minimum weight cycle in a graph? by user1696230
I think you are pretty close. Once you do FW loop in O(|V|³), you have the weight of a shortest path between each pair of vertices, let’s call it D(i, j). Now the solution to our main problem is as follows: Brute force for the vertex which is gonna be in a loop, say it’s X. Also brute force on Y, which is gonna be the last vertex in the cycle, before X itself. The weight of this cycle is obviously D(X, Y) + W(Y, X). Then you just choose the one with the smallest value. It’s obviously a correct answer, because: (1) All these values of D(X,Y)+D(Y,X) represent the weight of some(maybe non-simple) cycle; (2) If the optimal cycle is some a->…->b->a, then we consider it when X=a, Y=b;
To reconstruct the answer, first you need to keep track of which X, Y gave us the optimal result. Once we have this X and Y, the only thing remaining is to find the shortest path from X to Y, which can be done in various ways. by llaki
 mentions that both Dijkstra and Floyd Warshall algorithm can be used to find the minum weighted cycle in undirected graph.
 mentions 3 algorithms used to solve this problem. Except the Digkstra and Floyd Warshall algorithms, a brute force algorithm is also mentioned here. The complexity of each algorithm is analyzed here.
- brute force O(EV²)
- by using Dijkstra O(E*(V+E)log V) recall that the Dijkstra algorithm has the complexity of O((V+E)log V)
- by using Floyd Warshall O(V³)
 gives two ample codes used to solve the problem .
 gave a nice solution for problem 
using namespace std;
#define inf 1044266558
typedef struct Point
int x, y;
Point operator - ( const Point &b ) const
c.x = x-b.x; c.y = y-b.y;
double operator * ( const Point &b ) const
Point h, s;
int n, m, ans, road;
bool Jud(Point x, Point y, Point z)
if((x.x<z.x && y.x<z.x) || (x.y<z.y && y.y<z.y) || (x.x>z.x && y.x>z.x) || (x.y>z.y && y.y>z.y))
int i, j, k, flag;
memset(road, 62, sizeof(road));
scanf("%d%d", &h[i].x, &h[i].y);
scanf("%d%d", &s[i].x, &s[i].y);
flag = 1;
if((s[i]-s[j])*(s[i]-h[k])<0 || (s[i]-s[j])*(s[i]-h[k])==0 && Jud(s[i], s[j], h[k]))
flag = 0;
road[i][j] = 1;
ans = inf;
road[i][j] = min(road[i][j], road[i][k]+road[k][j]);
ans = min(ans, road[i][i]);