阿里3.29笔试

两道题，感觉都不难，一道简单计算，一道floyd，都能快速AC

1. 每人可以投K票，求保证获胜最小的K

   n = int(input())
   a = list(map(int, input().split()))
   sum = 0
   ans = 0
   for i in a:
    if i>ans:
        ans = i
    sum+=i
   while((ans*n-sum)<=sum):
    ans+=1
   print(ans)

2. 找能在d距离内到达其他城市数量最少的城市

   def floyd(n, adj):
    ans = adj
    for k in range(n):
        for i in range(n):
            for j in range(n):
                ans[i][j] = min(ans[i][j], ans[i][k]+ans[k][j])
    return ans
   T = int(input())
   for i in range(T):
    n,m,d = list(map(int, input().split()))
    adj_matrix = [[float('inf')]*n for i in range(n)]
    for j in range(n):
        adj_matrix[j][j] = 0
    for j in range(m):
        l,r,w = list(map(int, input().split()))
        adj_matrix[l][r] = w
        adj_matrix[r][l] = w
    dis = floyd(n, adj_matrix)
    ans = 0
    fewest = n
    for j in range(n):
        count = 0
        for to in dis[j]:
            if to<=d:count+=1
        if count<=fewest:
            fewest = count
            ans = j
    print(ans)