n,m = [int(x) for x in raw_input().split()]
use = range(500)
edge = []
for i in range(m):
v1,v2,ecost = [int(x) for x in raw_input().split()]
v1 -= 1
v2 -= 1
edge.append((v1,v2,ecost))
for i in range(len(use)):
use[i] = 0
use[0] = 1
cost = 0
for i in range(n-1):
mincost = (0,0,999999999)
for j in edge :
if use[j[0]]+use[j[1]] == 1 and mincost[2] > j[2] :
mincost = j
print mincost
cost += mincost[2]
if use[mincost[0]] == 0 :
use[mincost[0]] = 1
else :
use[mincost[1]] = 1
print cost
python实现的prim算法求MST O(MN)
最新推荐文章于 2024-07-14 20:56:17 发布