Find the minimum weight cycle in an undirected graph
Find the minimum weighted cycle in an undirected graph can be found from[1]. An example from [1] is shown here to well explain this problem.
2 + 6 + 2 + 4 = 14 [1]
Q&A from [2]
Q
“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
A
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
[3] mentions that both Dijkstra and Floyd Warshall algorithm can be used to find the minum weighted cycle in undirected graph.
[6] 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)
[5] gives two ample codes used to solve the problem [4].
[7] gave a nice solution for problem [8]
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define inf 1044266558
typedef struct Point
{
int x, y;
Point operator - ( const Point &b ) const
{
Point c;
c.x = x-b.x; c.y = y-b.y;
return c;
}
double operator * ( const Point &b ) const
{
return x*b.y-y*b.x;
}
}Point;
Point h[505], s[505];
int n, m, ans, road[505][505];
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))
return 1;
return 0;
}
int main(void)
{
int i, j, k, flag;
while(scanf("%d", &n)!=EOF)
{
memset(road, 62, sizeof(road));
for(i=1;i<=n;i++)
scanf("%d%d", &h[i].x, &h[i].y);
scanf("%d", &m);
for(i=1;i<=m;i++)
scanf("%d%d", &s[i].x, &s[i].y);
for(i=1;i<=m;i++)
{
for(j=1;j<=m;j++)
{
flag = 1;
for(k=1;k<=n;k++)
{
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;
break;
}
}
if(flag)
road[i][j] = 1;
}
}
ans = inf;
for(k=1;k<=m;k++)
{
for(i=1;i<=m;i++)
{
if(road[i][k]==inf)
continue;
for(j=1;j<=m;j++)
road[i][j] = min(road[i][j], road[i][k]+road[k][j]);
}
}
for(i=1;i<=m;i++)
ans = min(ans, road[i][i]);
if(ans>m)
printf("ToT\n");
else
printf("%d\n", m-ans);
}
return 0;
}Reference
[1]https://www.geeksforgeeks.org/find-minimum-weight-cycle-undirected-graph/
[2]https://stackoverflow.com/questions/22738785/minimum-weight-cycles-with-floyd-warshall-algorithm
[3]https://www.cnblogs.com/Yz81128/archive/2012/08/15/2640940.html
[4]http://poj.org/problem?id=1734
[5]https://blog.csdn.net/yo_bc/article/details/75042688
