//输入边集，返回最小生成树的权。
//复杂度为O(ElogE)，性能通常不如Prim。
func Kruskal(roads []graph.Edge, size int) (uint, error) {
if size < 2 || len(roads) < size-1 {
return 0, errors.New("illegal input")
}
graph.Sort(roads)
var list = make([]memo, size)
for i := 0; i < size; i++ {
list[i].cnt, list[i].group, list[i].next = 1, i, &list[i]
}
var active = &list[0]
var sum = uint(0)
for _, path := range roads {
if active.cnt == size {
break
}
var grpA, grpB = list[path.A].group, list[path.B].group
if grpA == grpB {
continue
}
sum += path.Dist
var another *memo
if list[grpA].cnt > list[grpB].cnt {
active, another = &list[grpA], &list[grpB]
} else {
active, another = &list[grpB], &list[grpA]
}
active.cnt += another.cnt
var tail = active.next
active.next, another.next = another.next, tail
for another = active.next; another != tail; another = another.next {
another.group = active.group
}
}
if active.cnt != size {
return sum, errors.New("isolated part exist")
}
return sum, nil
}
//输入边集，返回最小生成树的权。
//复杂度为O(ElogE)，性能通常不如Prim。
func Kruskal(roads []graph.Edge, size int) (sum uint, fail bool) {
if size < 2 || len(roads) < size-1 {
return 0, true
}
graph.Sort(roads)
var list = make([]memo, size)
for i := 0; i < size; i++ {
list[i].cnt, list[i].group, list[i].next = 1, i, &list[i]
}
var active = &list[0]
sum = uint(0)
for _, path := range roads {
if active.cnt == size {
break
}
var grpA, grpB = list[path.A].group, list[path.B].group
if grpA == grpB {
continue
}
sum += path.Dist
var another *memo
if list[grpA].cnt > list[grpB].cnt {
active, another = &list[grpA], &list[grpB]
} else {
active, another = &list[grpB], &list[grpA]
}
active.cnt += another.cnt
var tail = active.next
active.next, another.next = another.next, tail
for another = active.next; another != tail; another = another.next {
another.group = active.group
}
}
return sum, active.cnt != size
}
