可爱抱抱呀😥：``` dp[i][j][z] = min(dp[i][j][z], max(dp[i][v][z], dp[v][j][0])); dp[i][j][z] = min(dp[i][j][z], max(dp[i][v][0], dp[v][j][z])); if (z != 0) { dp[i][j][z] = min(dp[i][j][z], max(dp[i][v][z - 1], dist[v][j])); dp[i][j][z] = min(dp[i][j][z], max(dp[v][j][z - 1], dist[i][v])); } ``` 这段代码里，为啥更新k次传送门的数据，只用考虑[0,z]和[1,z-1]的组合？而不是所有和为z的次数组合？