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Ex - Make Q

Score : $625$ points

Problem Statement

There is a simple undirected graph with $N$ vertices and $M$ edges. The edges are initially painted white. The vertices are numbered $1$ through $N$, and the edges are numbered $1$ through $M$. Edge $i$ connects vertex $A_i$ and vertex $B_i$, and the cost required to paint it black is $C_i$.

"Making a Q" means painting four or more edges so that:

• all but one of the edges painted black form a simple cycle, and
• the edge painted black not forming the cycle connects a vertex on the cycle and another not on the cycle.

Determine if one can make a Q. If one can, find the minimum total cost required to make a Q.

Constraints

• $4\leq N \leq 300$
• $4\leq M \leq \frac{N(N-1)}{2}$
• $1 \leq A_i < B_i \leq N$
• $(A_i,B_i) \neq (A_j,B_j)$, if $i \neq j$.
• $1 \leq C_i \leq 10^5$
• All input values are integers.

Input

The input is given from Standard Input in the following format:

$N$ $M$

$A_1$ $B_1$ $C_1$

$A_2$ $B_2$ $C_2$

$\vdots$

$A_M$ $B_M$ $C_M$

Output

If one can make a Q, print the minimum total cost required to do so; otherwise, print $-1$.

5 6
1 2 6
2 3 4
1 3 5
2 4 3
4 5 2
3 5 1

15

By painting edges $2,3,4,5$, and $6$,

• edges $2,4,5$, and $6$ forms a simple cycle, and
• edge $3$ connects vertex $3$ (on the cycle) and vertex $1$ (not on the cycle),

so one can make a Q with a total cost of $4+5+3+2+1=15$. Making a Q in another way costs $15$ or greater, so the answer is $15$.

4 4
1 2 1
2 3 1
3 4 1
1 4 1

-1

6 15
2 6 48772
2 4 36426
1 6 94325
3 6 3497
2 3 60522
4 5 63982
4 6 4784
1 2 14575
5 6 68417
1 5 7775
3 4 33447
3 5 90629
1 4 47202
1 3 90081
2 5 79445

78154